Notes Prepared By: Mohammad Kamrul...

Notes Prepared By:

Mohammad Kamrul Arefin

Notes on Applied Mathematics

for Business and Economics Students

Table of Contents

Contents Page

Differential Calculus 1-64

Applications of Differential Calculus 65-81

Integral Calculus 82-108

Applications of Integral Calculus 109-127

Matrices and its Applications 128-142

Linear Programming & Simplex Method 143-177

Mohammad Kamrul Arefin, MSc. in Quantitative Finance, University of Glasgow Page | 1

Applied Mathematics

Practice Questions and Answers # 1

Problem Set 7-2:

Question No. 5:

2 2

1

1 1 1 0lim 0

1 1 1 2x

x

x

Question No. 6:

2 2

1

1 1 1 2lim 1

1 1 1 2x

x

x

Question No. 7:

2 2

2

5 2 5 9lim o r u n d e fin e d

2 2 2 0

A n s : L im it d o e s n o t e x is t

x

x

x

Question No. 9:

1 13 3lim ( 1) ( 1)

x a

x a

Question No. 10:

1 12 2

2 4

lim ( 1) ( 2 4 1) 2 5 5x

x


Question No. 13:

22

32

3 94 ( ) 9 4 ( ) 94 9 02 4lim

32 3 3 3 02 ( ) 32

It a p p e a rs f ro m th e a n s w e r th a t d e n o m in a to r a n d n u m e ra to r

h a v e a c o m m o n fa c to r a n d th e lim it c a n b e e v a lu a te d a f te r

e m p lo y in g th e p ro c e s s o f c a n c e lla t io n .

x

x

x

2 2 2

3 3 3 32 2 2 2

4 9 ( 2 ) 3 ( 2 3)( 2 3 )lim lim lim lim ( 2 3)

2 3 2 3 2 3

32 ( ) 3 3 3 62

A n s : 6

x x x x

x x x xx

x x x

Question No. 14:

22

53

5 2 59 ( ) 2 5 9 ( ) 2 59 2 5 03 9lim

5 53 5 03( ) 5 3( ) 53 3



e m p lo y in g th e p ro c e s s o f c a n c e

x

x

x

2 2 2

5 5 5 53 3 3 3

lla t io n .

9 2 5 (3 ) 5 (3 5 )(3 5 )lim lim lim lim (3 5 )

3 5 3 5 3 5

53( ) 5 5 5 1 03

A n s : 1 0

x x x x

x x x xx

x x x


Question No. 16:

3

40

3 3

4 30 0

0lim

0




1 1lim lim o

( ) 0

x

x x

x

x

x x

x x x x

r u n d e fin e d


Question No. 19:

0

0

( ) ( 2 3 ) (0 ){2 3(0 )} 0lim

0 0




( ) ( 2lim

h

h

h x h x

h

h x

0

3 )lim ( 2 3 ) {2 3(0 )} 2

A n s : 2 x

h

hx h x x

h

Question No. 23:

3 3

1

3

1 1

1 1 1 0lim

1 1 1 0

It a p p e a rs fro m th e a n s w e r th a t d e n o m in a to r a n d n u m e ra to r


e m p lo y in g th e p ro c e s s o f c a n c e lla tio n .

1 (lim lim

1

x

x x

x

x

x x

x

2

2

1

1)( 1)lim ( 1) 3

1

A n s : 3

x

x xx x

x


Question No. 24:

3 3

2

2 2 2 6lim o r u n d e fin e d

2 2 2 0


x

x

x

Question No. 25:

2 2 2

2 2lim ( ) lim ( ) lim (1) 1

2 2 2

A n s : 1

x x x

x x

x x x

Question No. 26:

2 2

1 1

2

1

1 1 0lim ( ) lim ( )

1 1 1 0

It a p p e a rs fro m th e a n s w e r th a t d e n o m in a to r a n d n u m e ra to r


e m p lo y in g th e p ro c e s s o f c a n c e lla tio n .

lim (

x x

x

x x

x x x

x

1 1

1 ( 1)( 1)) lim lim ( 1) 2

1 1

A n s : 2

x x

x xx

x x

Question No. 29:

2 2

2 22

2

22

6 2 2 6 0lim

4 2 4 0




6lim

4

x

x

x x

x

x x

x

2

2 2

2 2

3 2 6 ( 3 ) 2 ( 3 )lim lim

( 2 )( 2 ) ( 2 )( 2 )

( 3 )( 2 ) ( 3 ) ( 2 3 ) 5lim lim

( 2 )( 2 ) ( 2 ) ( 2 2 ) 4

5A n s :

4

x x

x x

x x x x x x

x x x x

x x x

x x x


Applied Mathematics


Problem Set 7-3:

Question No. 3:

2

2

2

( )1

T h e a b o v e fu n c tio n w ill b e c o n tin u o u s fo r a ll th e v a lu e s o f x e x c e p t w h e n ,

1 0

, 1

, 1 o r im a g in a ry n u m b e r

T h e re fo re w e c a n c o n c lu d e th a t th e a b o v e fu n c tio n h a v e n o d is c o n tin u ity .

xf x

x

x

o r x

o r x

Question No. 4:

2

2

2

2( )

4


4 0

, 4

, 2

T h e re fo re w e c a n c o n c lu d e th a t th e a b o v e fu n c tio n h a v e d is c o n tin u ity w h e n x = + 2 o r -2 .

xf x

x

x

o r x

o r x

Question No. 5:

24

( )2


2 0

, 2

T h e re fo re w e c a n c o n c lu d e th a t th e a b o v e fu n c tio n h a v e d is c o n tin u ity w h e n th e v a lu e o f x = 2 .

xf x

x

x

o r x


Question No. 6:

2

2

2

2( )

5 6


5 6 0

, 2 3 6 0

, ( 2 )( 3 ) 0

, ( 2 ) 0 o r ( 3 ) 0

. 2 , 3

T h e re fo re w e c a n c o n c lu d e th a t th e a b o

xf x

x x

x x

o r x x x

o r x x

e ith e r x x

i e x

v e fu n c tio n h a v e d is c o n tin u ity w h e n th e v a lu e o f x = 2 o r 3 .

Question No. 7:

1 1

1 1

( ) { 1; x > 1 }

( ) { ; x 1 }

b e c a u s e x > 1 a n d a ls o x 1,

w e n e e d to c h e c k if th e fu n c tio n is c o n t in u o u s o r d is c o n tin u o u s a t p o in t x = 1

lim ( 1) 2 a n d lim ( ) 1

, lim ( 1) lim ( )

T h e re fo re w e c

x x

x x

f x x

f x x

x x

h e re x x

a n c o n c lu d e th a t th e fu n c tio n is d is c o n t in u o u s fo r x = 1

Question No. 8:

1 1

1 1

( ) { 1; x 1 }

( ) {2 ; x < 1 }

b e c a u s e x 1 a n d a ls o x <1,


lim ( 1) 2 a n d lim ( 2 ) 2

, lim ( 1) lim ( )

T h e re fo re w e c

x x

x x

f x x

f x

x

h e re x x

a n c o n c lu d e th a t th e fu n c tio n h a v e n o d i s c o n tin u ity .


Question No. 9:

2

2 2

2 2

2

2 2

( ) { 1; x }

( ) { 3; x < 2 }

b e c a u s e x a n d a ls o x < 2 ,


lim ( 1) 2 + 1 = 5 a n d lim ( 3) 2 3 5

, lim ( 1) lim (

x x

x x

f x x

f x x

x x

h e re x

3)

T h e re fo re w e c a n c o n c lu d e th a t th e fu n c t io n h a v e n o d is c o n tin u ity .

x

Question No. 10:

2 2

2

( ) { 1; x }2

( ) { 2 ; x < 2 }

b e c a u s e x a n d a ls o x < 2 ,


2lim ( 1) ( 1) = 0 a n d lim ( 2 ) 2 2 0

2 2

, lim (

x x

x

xf x

f x x

xx

xh e re

2

1) lim ( 2 )2

T h e re fo re w e c a n c o n c lu d e th a t th e fu n c t io n h a v e n o d is c o n tin u ity .

x

x


Additional Question 1:

1

2

1

2

1 1( ) { ; 0 < x < }

2 2

1 1( ) { ; }

2 2

3 1( ) { ; < x < 1 }

2 2

1 1 1b e c a u s e 0 < x < , a n d a ls o < x < 1,

2 2 2

1w e n e e d to c h e c k if th e fu n c tio n is c o n t in u o u s o r d is c o n tin u o u s a t p o in t x =

2

1 1 1lim ( ) ( ) = 0 ,

2 2 2

lim (

x

x

f x x

f x x

f x x

x

x

1

2

1 1 1

2 2 2

1 1) a n d

2 2

3 3 1lim ( ) 1

2 2 2

1 1 3, lim ( ) lim ( ) lim ( )

2 2 2

1T h e re fo re w e c a n c o n c lu d e th a t th e fu n c t io n is d is c o n tin u o s a t

2

x

x x x

x

h e re x x

x


2

3

2 2

1

( ) { ; 0 < x < 1 }

( ) { ; 1 x < 2}

1( ) { ; 2 x < 3 }

4

b e c a u s e 0 < x < 1 , 1 x < 2 a n d a ls o 2 x < 3,

w e n e e d to c h e c k if th e fu n c tio n is c o n t in u o u s o r d is c o n tin u o u s

a t th e b re a k p o in t x = 1 a n d 2

lim ( ) (1 ) = 1 , limx x

f x x

f x x

f x x

x

3

1 1

2 3

1 1 1

2 2 3

2 2 2

2

2

1 1( ) 1 a n d lim ( )

4 4

1, lim ( ) lim ( ) lim ( )

4

T h e re fo re w e c a n c o n c lu d e th a t th e fu n c t io n is d is c o n tin u o s a t 1

1lim ( ) ( 2 ) = 4 , lim ( ) 2 a n d lim ( ) 2

4

, lim ( ) li

x

x x x

x x x

x

x x

h e re x x x

x

x x x

h e re x

3

2 2

1m ( ) lim ( )

4

T h e re fo re w e c a n c o n c lu d e th a t th e fu n c t io n is d is c o n tin u o s a t 2

x x

x x

x



2

1 1

( ) { ; 0 < x < 1}

( ) {2 ; 1 x 2 }

1( ) { ; x > 2}

2

b e c a u s e 0 < x < 1 , 1 x 2 a n d a ls o x > 2 ,

w e n e e d to c h e c k if th e fu n c t io n is c o n t in u o u s o r d is c o n tin u o u s

a t th e b re a k p o in t x = 1 a n d 2

lim ( ) (1) = 1 , lim (x x

f x x

f x x

f x x x

x

2 2

1

2

1 1 1

2 2

2

2

2 ) 2 1 1

1 1 1a n d lim ( ) 1 1

2 2 2

1, lim ( ) lim ( 2 ) lim ( )

2

T h e re fo re w e c a n c o n c lu d e th a t th e fu n c t io n is d is c o n tin u o u s a t 1

lim ( ) ( 2 ) = 2 , lim ( 2 ) 2 2 0

1a n d lim (

2

x

x x x

x x

x

x

x x

h e re x x x x

x

x x

x x

2

2

1 1 1

1) 2 2 0

2

1, lim ( ) lim ( 2 ) lim ( )

2


x x x

h e re x x x x

x



0

3( ) {3 2 ; < 0}

2

3( ) {3 2 ; 0 x < }

2

3( ) { 3 2 ; x }

2

3 3 3b e c a u s e < 0 , 0 x < a n d a ls o x ,

2 2 2

w e n e e d to c h e c k if th e fu n c tio n is c o n t in u o u s o r d is c o n tin u o u s

3a t th e b re a k p o in t x = 0 a n d

2

lim (3 2 )x

f x x x

f x x

f x x

x

x

0

0

0 0 0

3 3

2 2

3

2

3, lim (3 2 ) 3

a n d lim ( 3 2 ) 3

, lim (3 2 ) lim (3 2 ) lim ( 3 2 )


lim (3 2 ) 6 , lim (3 2 ) 0

a n d lim ( 3 2

x

x

x x x

x x

x

x

x

h e re x x x

x

x x

3 3

2 2

) 6

, lim (3 2 ) lim (3 2 ) lim ( 3 2 )

3T h e re fo re w e c a n c o n c lu d e th a t th e fu n c t io n is d is c o n tin u o s a t

2

x vx x

x

h e re x x x

x


Applied Mathematics


Problem Set 7-4:

Apply the three steps of the delta method to determine derivatives of x for each of the

following. Then find the slope of the curve of f(x) for the stated value of x.

Question No. 1:

0

0 0 0

( ) 3 2

{3( ) 2} {3 2}D e riv a t iv e o f x = ( ) lim

3 3 2 3 2 3lim lim lim 3 3

S lo p e a t x = 1 ; (1) 3

h

h h h

f x x

x h xf x

h

x h x h

h h

f

Question No. 2:

0

0 0 0

( ) 2 0 .5

{2 0 .5 ( )} {2 0 .5 }D e riv a t iv e o f x = ( ) lim

2 0 .5 0 .5 2 0 .5 0 .5lim lim lim ( 0 .5 ) 0 .5

S lo p e a t x = 1 ; (1) 0 .5

h

h h h

f x x

x h xf x

h

x h x h

h h

f

Question No. 3:

2

2 2

0

2 2 2 2

0 0

0 0

( ) 2 1

{ ( ) 2 ( ) 1} { 2 1}D e riv a tiv e o f x = ( ) lim

2 2 2 1 2 1 2 2lim lim

( 2 2 )lim lim ( 2 2 ) 2 2

S lo p e a t x = 1 ; (1) 2 (1) 2 0

h

h h

h h

f x x x

x h x h x xf x

h

x xh h x h x x xh h h

h h

h x hx h x

h

f


Question No. 4:

2

2 2

0

2 2 2 2

0 0

0 0

( ) 3 1 2 2

{3( ) 1 2 ( ) 2} {3 1 2 2}D e riv a tiv e o f x = ( ) lim

3 6 3 1 2 1 2 2 3 1 2 2 6 3 1 2lim lim

(6 3 1 2 )lim lim (6 3 1 2 ) 6 1 2

S lo p e a t x = 3 ; (3 ) 6 (3 )

h

h h

h h

f x x x

x h x h x xf x

h

x xh h x h x x xh h h

h h

h x hx h x

h

f

1 2 6

Question No. 5:

0

2

0 0

2 2 2 20 0

2

1( )

1 1

D e riv a tiv e o f x = ( ) lim

( )

( )lim lim

1 1 1 1lim { . } lim

1 1S lo p e a t x = 2 ; ( 2 )

2 4

h

h h

h h

f xx

x h xf x

h

x x h x x h

x x h x h x

h h

h

x h x h x h x x x

f


Question No. 6:

2

2 2

0

2 2 2 2 2

2 2 2 2 2

0 0

2 2 2 2

4 3 2 2 4 3 2 2

0 0

4 3 2 20

1( )

1 1

( )D e riv a tiv e o f x = ( ) lim

( ) ( 2 )

( ) ( 2 )lim lim

2 2

2 2lim lim

( 2 ) 1lim { .

2

h

h h

h h

h

f xx

x h xf x

h

x x h x x xh h

x x h x x xh h

h h

x x xh h xh h

x x h x h x x h x h

h h

h x h

x x h x h h

4 3 2 2 4 3

0

3

2 2 2} lim

2

2 2S lo p e a t x = 1; ( 1) 2

( 1) ( 1)

h

x h x

x x h x h x x

f

Question No. 7:

3

3 3

0

3 2 2 3 3 2 2 3

0 0

2 2

2 2 2

0 0

2

( ) 2

{ ( ) 2 ( )} { 2 }D e riv a tiv e o f x = ( ) lim

3 3 2 2 2 3 3 2lim lim

(3 3 2 )lim lim (3 3 2 ) 3 2

S lo p e a t x = 2 ; ( 2 ) 3 ( 2 ) 2

h

h h

h h

f x x x

x h x h x xf x

h

x x h x h h x h x x x h x h h h

h h

h x x h hx x h h x

h

f

1 4


Additional Question No.1:

3

3 3

0

3 2 2 3 3 2 2 3

0 0

2 2

2 2 2

0 0

( )

( )D e r iv a t iv e o f x = ( ) lim

3 3 3 3lim lim

(3 3 )lim lim (3 3 ) 3

h

h h

h h

f x x

x h xf x

h

x x h x h h x x h x h h

h h

h x x h hx x h h x

h


( )

0

0 0 0

2 3

0

2 3

0

( )

D e riv a tiv e o f x = ( ) lim

. ( 1) ( 1)lim lim lim

1lim { ( 1)} ; [N o te : {1 ... . . . . . . . . . . .

1

1lim {1 ... . . . . . . . . .

1

x

x h x

h

x h x x h h

x

h h h

x h x

h

x

h

f x e

e ef x

h

e e e e e ee

h h h

x x xe e e

h

h h he

h

2 3

0

2

0

2

0

. . 1}

1lim { ... . . . . . . . . . . . }

1

1 1lim { ( ... . . . . . . . . . . . }

1

lim {1 ... . . . . . . . . . . . }

x

h

x

h

x

h

x

h h he

h

h he h

h

h he

e



( )

0

0 0 0

lo g

lo g lo g .1

0

lo g 2 3

0

( )

D e r iv a t iv e o f x = ( ) lim

. ( 1) ( 1)lim lim lim

( 1)lim ; [N o te : ]

( 1)lim ; [N o te : {1 ...

1

ha

e ha e

e e

a

e

x

x h x

h

x h x x h h

x

h h h

x h h h

h

h

x x

h

f x a

a af x

h

a a a a a aa

h h h

ea e a a a

h

e x x xa e

h

2 2 3 3

0

2 2 3 3

0

2 2 3

0

. . . . . . . . . . .

1 lo g ( lo g ) ( lo g )lim {1 ... . . . . . . . . . . . 1}

1

1 lo g ( lo g ) ( lo g )lim { ... . . . . . . . . . . . }

1

1 lo g ( lo g ) ( lo g )lim [ { ... . . . . . . . . . . }

1

x

h

x

h

x

h

h a h a h aa

h

h a h a h aa

h

a h a h aa h

h

2 2 3

0

]

( lo g ) ( lo g )lim { lo g ... . . . . . . . . . . }

lo g

x

h

x

h a h aa a

a a



0

0

2 3 4

0

2 3 4

0

0

( ) lo g

lo g ( ) lo gD e riv a t iv e o f x = ( ) lim

1lim . lo g

1 1 1 1lim . lo g (1 ) ; [N o te : lo g (1 ) .. . . . . . . . . ]

2 3 4

1 1 1 1lim .{ ( ) ( ) ( ) .. . . . . . . . .

2 3 4

1lim . {

e

h

h

h

h

h

f x x

x h xf x

h

x h

h x

hx x x x x

h x

h h h h

h x x x x

hh

2 3

2 3 4

2 3

2 3 40

1 1 1 1... . . . . . . .

2 3 4

1 1 1 1lim { ... . . . . . . .

2 3 4

1

h

h h h

x x x x

h h h

x x x x

x


( ) lo g lo g . lo g

D e riv a tiv e o f x = ( ) ( lo g . lo g )

lo g (lo g )

1lo g

x e

e

e

e

e

e

e

f x a a x

df x a x

d x

da x

d x

ax


Applied Mathematics


Problem Set 7-5:

F in d ( ). D o n o t leave n eg a tive ex p o n en ts in an sw ers .f x

Question No. 1:

( ) 2

D e riv a tiv e o f f (x ) = ( ) [ ( ) ] [ 2 ]

( 2 ) ( ) 0

f x k

d df x f x k

d x d x

d dk

d x d x

Question No. 2:

( )

D e riv a tiv e o f f(x ) = ( ) [ ( )] [ ] 1

f x x

d df x f x x

d x d x

Question No. 3:

( ) / 2

D e r iv a t iv e o f f (x ) = ( ) [ ( ) ] [ / 2 ]

1 1 1 1[ . ] [ ] .1

2 2 2 2

f x x

d df x f x x

d x d x

d dx x

d x d x

Question No. 4:

( ) 2 3

D e riv a t iv e o f f (x ) = ( ) [ ( ) ] [ 2 3 ]

( 2 ) (3 ) 2 ( ) 0 2

f x x

d df x f x x

d x d x

d d dx x

d x d x d x


Question No. 5:

( ) / 3 4

D e riv a tiv e o f f (x ) = ( ) [ ( ) ] [ / 3 4 ]

1 1 1( ) ( 4 ) ( ) 0 .1

3 3 3 3

f x x

d df x f x x

d x d x

d x d dx

d x d x d x

Question No. 6:

( ) 2 / 3 3 ( / 2 ) 2 / 3 3 / 2

D e riv a t iv e o f f (x ) = ( ) [ ( ) ] [ 2 / 3 3 / 2 ]

3 3 3( 2 / 3 ) (3 / 2 ) 0 ( ) ( )

2 2 2

f x x x

d df x f x x

d x d x

d d d dx x x

d x d x d x d x

Question No. 7:

2

2

2 2

2 1

( ) 3 2 5

D e riv a t iv e o f f (x ) = ( ) [ ( ) ] [3 2 5 ]

(3 ) ( 2 ) (5 ) 3 ( ) 2 ( ) 0

3 ( 2 ) 2 (1) 6 2

f x x x

d df x f x x x

d x d x

d d d d dx x x x

d x d x d x d x d x

x x

Question No. 8:

3 2

3 2

3 2

3 2

3 1 2 1 2

( ) 1 23 2

D e riv a tiv e o f f (x ) = ( ) [ ( ) ] [ 1 2 ]3 2

( ) ( ) ( ) (1 2 )3 2

1 1( ) ( ) 1 0

3 2

1 1(3 . ) ( 2 . ) 1 1

3 2

x xf x x

d d x xf x f x x

d x d x

d x d x d dx

d x d x d x d x

d dx x

d x d x

x x x x


Question No. 12:

2

2

2

2

2 1

( )

D e riv a t iv e o f f (x ) = ( ) [ ( ) ] [ ]

( ) ( ) ( )

( ) ( ) 0

( 2 . ) 2

f x a x b x c

d df x f x a x b x c

d x d x

d d da x b x c

d x d x d x

d da x b x

d x d x

a x b a x b

Question No. 20:

2

2

2

(1 0 4 7 )

(1 0 ) ( 4 ) (7 )

1 0 ( ) 4 ( ) 0 1 0 ( 2 ) 0 0 2 0

dy x

d y

d d dy x

d y d y d y

d dy x y y

d y d y

Question No. 22:

2 3

2 3

(3 2 )

(3 ) ( 2 )

0 0 0

dp w p

d x

d dp w p

d x d x

Question No. 24:

2 3

2 3 2

1 1 1 2

2

( 3 5 )

1( ) (3 ) (5 ) ( ) 3 ( ) 0

( ) 3 ( 2 ) ( 1 ) 6 6 6

d am a

d m m

d a d d d dm a a m

d m m d m d m d m m d m

d aa m m a m m a m m m

d m m


Find the slope of the tangent to each of the following curves at the indicated value

of x:

Question No. 25:

( ) 3

( ) [ ( ) ] [3 ] 0

a t x = 1 ; (1) 0

f x

d dS lo p e f x f x

d x d x

S lo p e f

Question No. 29:

2

2

2

( ) 3 2 5

( ) [ ( )] [3 2 5 ]

(3 ) ( 2 ) (5 ) 3 ( 2 ) 2 (1) 0 6 2

a t x = 0 .5 ; (0 .5 ) 6 (0 .5 ) 2 1

f x x x

d dS lo p e f x f x x x

d x d x

d d dx x x x

d x d x d x

S lo p e f

Question No. 30:

2

2

2

( ) 3 2 5

( ) [ ( )] [3 2 5 ]

(3 ) ( 2 ) (5 ) 3 ( 2 ) 2 (1) 0 6 2

a t x = 0 .5 ; (0 .5 ) 6 (0 .5 ) 2 1

f x x x


d x d x

d d dx x x x

d x d x d x

S lo p e f


Question No. 33:

1 / 3

1 / 3

1 21

1 / 3 3 3

23 2

3

2 2 23 .

33 3 3

2

( ) 8

( ) [ ( )] [8 ]

1 8 8 1 8 1(8 ) ( ) 8 ( ) 1 1 1 1

3 3 3 3

8 1 8 1 8 1 a t x = 8 ; (8 ) 1 1 1

3 3 3(8 ) ( 2 ) 2

8 1 8 1 2 51 1 1

3 2 3 4 3 3

f x x x


d x d x

d dx x x x

d x d x xx

S lo p e f

Find the value of x for which the slope is 0

Question No. 35:

2

2

2

( ) 1 0 3 3

( ) [ ( ) ] [1 0 3 3 ]

(1 0 ) (3 ) (3 ) 0 3 ( 2 ) 3 (1) 6 3

0

, 6 3 0

, 6 3

, 1 / 2

S lo p e w ill b e 0 w h e n x = 1 /2

f x x x


d x d x

d d dx x x x

d x d x d x

S lo p e

o r x

o r x

o r x


Question No. 37:

1 / 3

1 / 3

1 21

1 / 3 3 3

2

3

2

3

2

3

2

3

2.3

33

2

( ) 3 4 ; x > 0

( ) [ ( ) ] [3 4 ]

1 1(3 ) ( 4 ) 3 ( ) 4 (1) 4 4

3

0

1, 4 0

1, 4

1,

4

1, ( )

4

1,

6 4

1,

8

a s x > 0 , h e n c e s lo p e is 0 w h e n

f x x x


d x d x

d dx x x x

d x d xx

S lo p e

o r

x

o r

x

o r x

o r x

o r x

o r x

x = 1 /8


Question No. 41: If the total cost of producing y yards of Yardall is,

C(y) = 0.001y2+2y+500, Find the marginal cost at outputs of a) 1,000 yards b)2,000 yards

2

2

M a rg in a l C o s t ( ) [ ( ) ] [0 .0 0 1 2 5 0 0 ]

(0 .0 0 1 ) ( 2 ) (5 0 0 ) 0 .0 0 1( 2 ) 2 (1) 0 0 .0 0 2 2

) M a rg in a l C o s t a t o u tp u ts = 1 0 0 0 ya rd s

(1 0 0 0 ) 0 .0 0 2 (1 0 0 0 ) 2 $ 4 p e r ya rd

) M a rg in a l C o s t a t o u t

d dC y C y y y

d y d y

d d dy y y y

d y d y d y

a

C

b

p u ts = 2 0 0 0 ya rd s

( 2 0 0 0 ) 0 .0 0 2 ( 2 0 0 0 ) 2 $ 6 p e r ya rdC

Question No. 41: If the total cost of producing t tons of Tonal is,

C(t) = 0.0005t3-0.3t

2+100t+30,000,

Find the marginal cost at outputs of a) 100 tons b)200 tons

3 2

3 2

2 2

M a rg in a l C o s t ( ) [ ( ) ] [ 0 .0 0 0 5 0 .3 1 0 0 3 0 0 0 0 ]

(0 .0 0 0 5 ) ( 0 .3 ) (1 0 0 ) (3 0 0 0 0 )

0 .0 0 0 5 (3 ) 0 .3 ( 2 ) 1 0 0 (1) 0 0 .0 0 1 5 0 .6 1 0 0

) M a rg in a l C o s t a t o u tp u ts = 1 0 0 to n s

(1 0 0 ) 0 .0

d dC t C t t t t

d t d t

d d d dt t t

d t d t d t d t

t t t t

a

C

2

2

0 1 5 (1 0 0 ) 0 .6 (1 0 0 ) 1 0 0 $ 5 5 p e r to n

) M a rg in a l C o s t a t o u tp u ts = 2 0 0 to n s

( 2 0 0 ) 0 .0 0 1 5 ( 2 0 0 ) 0 .6 ( 2 0 0 ) 1 0 0 $ 4 0 p e r to n

b

C


Applied Mathematics


Problem Set 7-6:


Question No. 1:

5

5 5 1

4

4 4 4

( ) (6 5 )

( ) [ ( ) ] [ (6 5 ) ] 5 (6 5 ) (6 5 )

5 (6 5 ) [ (6 ) (5 )]

5 (6 5 ) [6 ( ) 0 ] 5 (6 5 ) [6 ] 3 0 (6 5 )

f x x

d d df x f x x x x

d x d x d x

d dx x

d x d x

dx x x x

d x

Question No. 2:

5

5 5 1

4

4 4

( ) ( 2 6 )

( ) [ ( ) ] [ ( 2 6 ) ] 5 ( 2 6 ) ( 2 6 )

5 ( 2 6 ) [ ( 2 ) (6 )]

5 ( 2 6 ) [ 2 ] 1 0 ( 2 6 )

f x x

d d df x f x x x x

d x d x d x

d dx x

d x d x

x x

Question No. 3:

3

3 3 1

2 2

( ) ( 2 )

( ) [ ( ) ] [ ( 2 ) ] 3 ( 2 ) ( 2 )

3 ( 2 ) ( 2 ) 2 4

f x x

d d df x f x x x x

d x d x d x

x x


Question No. 4:

3

3 3

3

3

1

1 11

3

2

2323

( ) (6 )

1 ( ) [ ( ) ] [ (6 ) ] (6 ) (6 )

1 1 2(6 ) (6 ) 2

(6 )(6 )

f x x

d d df x f x x x x

d x d x d x

x

xx

Question No. 5:

1

2

1 11

2 2

1

2

1

2

( ) ( 4 )

1 ( ) [ ( )] [( 4 ) ] ( 4 ) ( 4 )

2

1 1 2 2 1( 4 ) ( 4 ) 2

2 ( 4 ) 2( 4 )

f x x

d d df x f x x x x

d x d x d x

x

x x xx

Question No. 6:

4

3

4 41

3 3

4 11

3 3

( ) (9 )

4 ( ) [ ( ) ] [ (9 ) ] (9 ) (9 )

3

4(9 ) (9 ) 1 2 (9 )

3

f x x

d d df x f x x x x

d x d x d x

x x

Question No. 7:

3

2

3 31

2 2

3 1 11

2 2 2

( ) (8 3 )

3 ( ) [ ( ) ] [ (8 3 ) ] (8 3 ) (8 3 )

2

3 3(8 3) { (8 ) (3 )} (8 3 ) (8 ) 1 2 (8 3) 1 2 8 3

2 2

f x x

d d df x f x x x x

d x d x d x

d dx x x x x

d x d x


Question No. 8:

5

3

5 51

3 3

2 2

233 3

( ) (1 2 9 )

5 ( ) [ ( ) ] [ (1 2 9 ) ] (1 2 9 ) (1 2 9 )

3

5(1 2 9 ) (1 2 ) 2 0 (1 2 9 ) 2 0 (1 2 9 )

3

f x x

d d df x f x x x x

d x d x d x

x x x

Question No. 9:

5

2 2

5

2 2

51

2 22

3 3

2 22 2

3

2 2 32

( ) (3 6 2 )

( ) [ ( ) ] [ (3 6 2 ) ]

5(3 6 2 ) (3 6 2 )

2

5 5(3 6 2 ) (6 6 ) (3 6 2 ) 6 ( 1)

2 2

1 5 ( 1)(3 6 2 ) 1 5 ( 1) (3 6 2 )

f x x x

d df x f x x x

d x d x

dx x x x

d x

x x x x x x

x x x x x x

Question No. 10:

4

3 2 3

4

3 2 3

41

3 2 3 23

1

3 2 23

1 1

3 2 2 2 3 23 3

( ) ( 3 6 )

( ) [ ( ) ] [ ( 3 6 ) ]

4( 3 6 ) ( 3 6 )

3

4( 3 6 ) (3 6 6 )

3

4( 3 6 ) 3 ( 2 2 ) 4 ( 2 2 )( 3 6 )

3

f x x x x

d df x f x x x x

d x d x

dx x x x x x

d x

x x x x x

x x x x x x x x x x


Question No. 11:

1

2

1

2

11

2

1 1

2 2

1

2

( ) ( 2 3 )

( ) [ ( ) ] [ ( 2 3 ) ]

1( 2 3 ) ( 2 3 )

2

1 1 1( 2 3) ( 2 ) ( 2 3 )

2 ( 2 3 )( 2 3 )

f x x

d df x f x x

d x d x

dx x

d x

x x

xx

Question No. 12:

2

2 3

2

2 3

21

2 23

1 1

2 23 3

1

2 3

( ) (3 5 )

( ) [ ( ) ] [ (3 5 ) ]

2(3 5 ) (3 5 )

3

2 4(3 5 ) (6 ) 4 (3 5 )

3(3 5 )

f x x

d df x f x x

d x d x

dx x

d x

xx x x x

x


Question No. 13:

1

1 1

1 1 2

2

2

2

4 1( ) 4 4 ( 2 3 )

2 3 2 3

( ) [ ( ) ] [ 4 ( 2 3 ) ] 4 [( 2 3 ) ]

4 ( 1)( 2 3 ) ( 2 3 ) 4 ( 2 3 ) ( 2 )

88 ( 2 3 )

( 2 3 )

A n o th e r M e th o d

( 2 3 ) ( 4 ) 4 ( 2 3 )4

( ) ( )2 3 ( 2 3 )

( 2 3

f x xx x

d d df x f x x x

d x d x d x

dx x x

d x

xx

d dx x

d d x d xf x

d x x x

x

2 2

) (0 ) 4 ( 2 ) 8

( 2 3 ) ( 2 3 )x x

Do the following at your home:

Question No. 14:

6( )

3 5f x

x

Question No. 17: 2

9( )

(3 5 )f x

x

Question No. 18: 3

1 2( )

( 2 1 0 )f x

x

Question No. 15:

2

2 2 1

1 1 1 2

2

2

1( ) ( 2 )

1 1 1( ) [ ( 2 ) ] 2 ( 2 ) ( 2 )

1 1 12 ( 2 ) ( 2 ) 2 ( 2 )( 0 ) 2 ( 2 )( )

1 1 12 ( 2 )( ) 2 ( 2 )

f xx

d df x

d x x x d x x

dx x x

x d x x x

xx x x


Question No. 16:

3

2

3 3 1

2 2 2

2 2 2 2 1

2 2

2 3 2

2 3 2

1( ) (5 )

1 1 1( ) [ (5 ) ] 3 (5 ) (5 )

1 13 (5 ) (5 ) 3 (5 ) {0 ( 2 ) )

1 1 13 (5 ) ( 2 ) 6 (5 )

f xx

d df x

d x x x d x x

dx x

x d x x

xx x x

Question No. 19:

2 2

2

1 0( ) 5

(3 2 )

1 0 1 0( ) [5 ] (5 ) { }

(3 2 ) (3 2 )

(3 2 ) (1 0 ) 1 0 (3 2 )0 1 0 (3)

5 5(3 2 ) (3 2 )

3 05

(3 2 )

f x xx

d d df x x x

d x x d x d x x

d dx x

d x d x

x x

x


Question No. 20:

5( ) 0 .1

5 0 .2f x x

x

Question No. 21: 1

2

1( ) 3

(5 2 )

f x x

x

Question No. 22:

1

2

1( ) 2

(3 7 )

f x x

x


Question No. 23:

1 / 2

1 / 2 1 / 2

1 1

2 2

1

22

11

2

1

2

1 8F in d ( 2 ) i f ( ) 1 0

(5 2 )

1 8 1 8( ) [1 0 ] (1 0 ) { }

(5 2 ) (5 2 )

(5 2 ) (1 8 ) 1 8 { (5 2 ) }

1 0

{ (5 2 ) }

10 1 8 (5 2 ) (5 2 )

21 0

(5 2 )

11 8 . (5 2 ) ( 2 )

21 0

(5 2 )

1 81 0

g g x xx

d d dg x x x

d x x d x d x x

d dx x

d x d x

x

dx x

d x

x

x

x

1 3

2 2

3

2

2

3 3 3

22 2 2

2

3

1 1 81 0

(5 2 )(5 2 ) (5 2 )

1 8( ) 1 0

(5 2 )

1 8 1 8 3 .2( 2 ) 1 0 1 0 1 0

(5 4 ) (9 ) (3 )

3 .2 2 2 81 0 1 0

3 3 3

xx x

g x

x

g


Question No. 24: 2 1 / 3

4F in d (3 ) i f ( ) 7

( 1)h h x x

x


Applied Mathematics


Problem Set 7-7:


Question No. 1:

( ) (3 2 )( 2 5 )

( ) [ (3 2 )( 2 5 )] (3 2 ) ( 2 5 ) ( 2 5 ) (3 2 )

(3 2 )( 2 ) ( 2 5 )(3 ) 6 4 6 1 5 1 2 1 1

f x x x

d d df x x x x x x x

d x d x d x

x x x x x

Question No. 2:

( ) ( 7 3 )( 4 3 )

( ) [ ( 7 3 )( 4 3 )] (7 3 ) ( 4 3 ) ( 4 3 ) (7 3 )

(7 3 )( 3 ) ( 4 3 )(7 ) 2 1 9 2 8 2 1 1 9 4 2

f x x x


d x d x d x

x x x x x

Question No. 3:

2

2 2 2

2 2 2 2

( ) ( 2 )(3 5 )

( ) [( 2 )(3 5 )] ( 2 ) (3 5 ) (3 5 ) ( 2 )

( 2 )(3 ) (3 5 )( 2 ) 3 6 6 1 0 9 1 0 6

f x x x


d x d x d x

x x x x x x x x

Question No. 4:

2

2 2 2

2 2 2 2

( ) (3 )(5 6 )

( ) [(3 )(5 6 )] (3 ) (5 6 ) (5 6 ) (3 )

(3 )(5 ) (5 6 )( 2 ) 1 5 5 1 0 1 2 3(5 4 5 )

f x x x


d x d x d x

x x x x x x x x


Question No. 5:

4

4 4 4

3 4

3 4 3 3

( ) ( 1)

( ) [ ( 1) ] ( 1) ( 1) ( )

4 ( 1) ( 1) ( 1) (1)

4 ( 1) ( 1) ( 1) ( 4 1) ( 1) (5 1)

f x x x


d x d x d x

dx x x x

d x

x x x x x x x x

Do the following at your home: Question No. 6: 2 3

( ) ( 5 )f x x x

Question No. 7:

3

2 2( ) ( 3)f x x x Question No.8:

2

3 3( ) (6 1)f x x x

Question No.9:

1

2 3

1 1 1

2 2 23 3 3

1 11

2 2 23 3

1 11

2 23 3

1 1

2 2 1 23 3

1

2 3

2

2

( ) 2 (3 7 )

( ) [ 2 (3 7 ) ] 2 (3 7 ) (3 7 ) ( 2 )

12 { (3 7 ) (3 7 ) (3 7 ) ( 2 )

3

2{ (3 7 ) (6 )} 2 (3 7 )

3

2(6 ){ (3 7 ) (3 7 ) } 2 (3 7 )

3

(3 7 )4 {

(3

f x x x


d x d x d x

dx x x x

d x

x x x x

x x x x x

xx

x

1 1 2

2 23 3

2

1 12 2 2

2 23 3

2 2

1 11

2 2 2 1 2 23 3

2 2

2 2 3

2

2 3

2} 2 (3 7 ) 2 (3 7 ) { 1}

7 ) (3 7 )

2 3 7 5 72 (3 7 ) { } 2 (3 7 ) { }

(3 7 ) (3 7 )

2 (5 7 )(3 7 ) (3 7 ) 2 (5 7 )(3 7 )

2 (5 7 )2 (5 7 )(3 7 )

(3 7 )

xx x

x

x x xx x

x x

x x x x x

xx x

x


Do this your home: Question No. 10:

1

3 2( ) 3 ( 2 5 )f x x x

Question No. 11:

2

2 2 2

( )1

( 1) ( ) ( 1)

( ) [ ]1 ( 1)

( 1)(1) (1) 1 1

( 1) ( 1) ( 1)

xf x

x

d dx x x x

d x d x d xf x

d x x x

x x x x

x x x

Question No. 12:

2

2 2 2

2( )

1

( 1) ( 2 ) ( 2 ) ( 1)2

( ) [ ]1 ( 1)

( 1)(1) ( 2 )(1) 1 2 3

( 1) ( 1) ( 1)

xf x

x

d dx x x x

d x d x d xf x

d x x x

x x x x

x x x

Do these at your home: Question No. 13:

2

( )2 3

xf x

x

Question No. 14:

3

( )3 5

xf x

x

Question No. 15:

2( )

3 2

xf x

x

Question No. 16:

2

3( )

1 2

xf x

x


Question No. 17:

1

2

1 1

2 2

1 1

22 2

1 11

2 2

1

21 1

12 2

2

1

2

( )

(3 2 )

(3 2 ) ( ) ( ) (3 2 )

( ) [ ]

(3 2 ) { (3 2 ) }

1(3 2 ) (1) ( ) (3 2 ) (3 2 )

2

(3 2 )

3(3 2 )

(3 2 ) (3 2 ) (3 )2 (3 2 )2

(3 2 ) (3 2 )

2 (3 2 ) 3

2 (3 2 )

(3

xf x

x

d dx x x x

d x d x d xf x

d xx x

dx x x x

d x

x

xxx

x xx

x x

x x

x

x

1 1 3

12 2 2

6 4 3 3 4 3 4

2 )2 (3 2 ) (3 2 ) 2 (3 2 ) 2 (3 2 )

x x x x

x x x x


Question No. 18:

1

2

1 1

2 2

1 1

22 2

1 11

2 2

1

2

1 1 1

2 2 2

1

2( )

( 2 3 )

( 2 3 ) ( 2 ) ( 2 ) ( 2 3 )2

( ) [ ]

( 2 3 ) { ( 2 3 ) }

1( 2 3 ) ( 2 ) ( 2 ) ( 2 3 ) ( 2 3 )

2

( 2 3 )

12 ( 2 3 ) 2

2 ( 2 3 ) ( 2 3 ) ( 2 ) ( 2 3 )

( 2 3 ) ( 2 3 )

2 ( 2 3 ) 2

( 2 3 )

xf x

x

d dx x x x

d x d x d xf x

d xx x

dx x x x

d x

x

x x

x x x x

x x

x x

x

1 1

2 2 2

1

2

3

2

4 6 2 2 6

2 ( 3 )( 2 3 ) ( 2 3 )

( 2 3 ) ( 2 3 ) ( 2 3 )( 2 3 )( 2 3 )

2 ( 3 )

( 2 3 )

x x x

xx x

x x xx x

x

x


1

2

1

2

1

2

A n o th e r M e th o d

2L e t Y =

( 2 3 )

2 lo g lo g { } ; [ ta k in g lo g in b o th s id e ]

( 2 3 )

lo g lo g ( 2 ) lo g { ( 2 3 ) }

1 lo g lo g ( 2 ) lo g ( 2 3 )

2

1 1 1 1 ( 2 ) ( 2 3 )

2 2 ( 2 3 )

1 1 1 1 ( 2 )

2 2 ( 2 3

x

x

xo r Y

x

o r Y x x

o r Y x x

d y d do r x x

Y d x x d x x d x

d yo r

Y d x x x

1 1

2 2

1 1

2 2

1 3

2 2

( 2 ))

1 1 1

( 2 3 )

2 3 ( )

( 2 3 )

2 2 3 ( ) ( )

( 2 3 )( 2 3 ) ( 2 3 )

2 2 ( 3 ) ( )

( 2 3 ) ( 2 3 )( 2 3 )

2 2 ( 3 ) ( )

( 2 3 ) ( 2 3 )

d yo r

Y d x x x

d y x xo r Y

d x x x

d x x xo r

d x x xx x

d x xo r

d xx x x

d x xo r

d xx x

Do these: Question No. 19:

2 1 / 3

2 1( )

( 5 )

xf x

x

20:

3 1 / 3

3 5( )

( 2 )

xf x

x


Additional Questions:

1 / 4

1 / 2 1 / 5

1 / 4

1 / 2 1 / 5

1 / 4

1 / 2 1 / 5

1 / 4 1 / 2 1 / 5

1 / 4

( 4 1)( )

( 2 3 ) (5 1)

( 4 1)L e t Y =

( 2 3 ) (5 1)

( 4 1) lo g lo g { } ; [ ta k in g lo g in b o th s id e ]

( 2 3 ) (5 1)

lo g lo g ( 4 1) lo g { ( 2 3 ) (5 1) }

lo g lo g ( 4 1) lo g

xf x

x x

x

x x

xo r Y

x x

o r Y x x x

o r Y x

1 / 2 1 / 5

( 2 3 ) lo g (5 1)

1 1 1 lo g lo g ( 4 1) lo g ( 2 3 ) lo g (5 1)

4 2 5

1 1 1 1 1 1 1 ( 4 1) ( 2 3 ) (5 1)

4 ( 4 1) 2 ( 2 3 ) 5 (5 1)

1 1 1 1 1 1 1 ( 4 ) ( 2 ) (5 )

4 ( 4 1) 2 ( 2 3 ) 5 (5 1)

1 1 1

( 4 1) ( 2

x x

o r Y x x x

d y d d do r x x x

Y d x x d x x d x x d x

d yo r

Y d x x x x

d yo r

Y d x x

1 / 4

1 / 2 1 / 5

1 / 4 2 2 2

1 / 2 1 / 5

1

3 ) (5 1)

1 1 1 { }

( 4 1) ( 2 3 ) (5 1)

( 4 1) ( 2 3 )(5 1) ( 4 1)(5 1) ( 4 1)( 2 3 ) { }

( 2 3 ) (5 1) ( 4 1)( 2 3 )(5 1)

( 4 1) 1 0 1 3 3 2 0 1 8 1 {

( 2 3 ) (5 1)

x x

d yo r Y

d x x x x

d y x x x x x x xo r

d x x x x x x

d y x x x x x xo r

d x x x

1 / 4 2

1 / 2 1 / 5

2

3 / 2 6 / 5 3 / 4

1 / 4 2

1 / 2 1 / 5 3 / 2 6

4 3}

( 4 1)( 2 3 )(5 1)

( 4 1) 1 8 2 5 { }

( 2 3 ) (5 1) ( 4 1)( 2 3 )(5 1)

(1 8 2 5 )

( 2 3 ) (5 1) ( 4 1)

( 4 1) (1 8 2 5 ). . { }

( 2 3 ) (5 1) ( 2 3 ) (5 1)

x

x x x

d y x x xo r

d x x x x x x

d y x xo r

d x x x x

d x x xi e

d x x x x x

/ 5 3 / 4

( 4 1)x


Applied Mathematics

Additional Problems from Differentiation

1

3 2 3

1 11

3 2 3 2 3 23 3

2

3 2 3 23

2

2

3 2 3

2

2

3 2 3

2

2

3 2 3

Q .1) ( ) ( 3 5 )

1 ( ) [ ( 3 5 ) ] ( 3 5 ) ( 3 5 )

3

1( 3 5 ) { ( ) 3 ( ) (5 )}

3

1 1. {3 3 .2 0}

3( 3 5 )

1 1. .3 ( 2 )

3( 3 5 )

( 2 )

( 3 5 )

f x x x

d df x x x x x x x

d x d x

d d dx x x x

d x d x d x

x x

x x

x x

x x

x x

x x

1 1 1

2

2

1Q .2 ) ( )

4 (8 6 )

1 1 1 ( ) [ ] { }

4 (8 6 ) 4 (8 6 )

1 1(8 6 ) ( 1)(8 6 ) (8 6 )

4 4

1 2(8 6 ) (8 )

4 (8 6 )

f xx

d df x

d x x d x x

d dx x x

d x d x

xx


3

2

3

2

3

2

31

2

5

2

5

2

5

2Q .3 ) ( )

2 7 (1 0 3 )

2 2 ( ) [ ] { (1 0 3 ) }

2 72 7 (1 0 3 )

2 3( )(1 0 3 ) (1 0 3 )

2 7 2

1 1 1(1 0 3 ) ( 3 ) .

9 3(1 0 3 )

1

3 (1 0 3 )

f x

x

d df x x

d x d xx

dx x

d x

x

x

x

1

3

1 1

3 3

1 11

3 3

4

3

4

3

43

1Q .4 ) ( )

(5 0 .3 )

1 1 ( ) [ ] ( ) { }

(5 0 .3 ) (5 0 .3 )

1( ) (5 0 .3 ) 1 ( )(5 0 .3 ) (5 0 .3 )

3

1 0 .3 11 (5 0 .3 ) ( 0 .3 ) 1 .

3 3(5 0 .3 )

11

1 0 (5 0 .3 )

f x x

x

d d df x x x

d x d x d xx x

d d dx x x x

d x d x d x

x

x

x


1

3 2

1 1

3 32 2

1 1

3 32 2

1 11

3 3 32 2

1 1

3 2 32 2

12

3 2

1

3 2

Q .5 ) ( ) 2 ( 5 )

( ) [ 2 ( 5 ) ] 2 [ ( 5 ) ]

2[ ( 5 ) ( 5 ) ( ) ]

12[ . ( 5 ) ( 5 ) ( 5 ) .1]

2

2[ ( 5 ) (3 5 ) ( 5 ) ]2

(3 5 )2 ( 5 )

( 5 )

f x x x x

d df x x x x x x x

d x d x

d dx x x x x x

d x d x

dx x x x x x x

d x

xx x x x x

xx x x

x x

13

3 2

1

3 2

3 3 3 3

1 1

3 32 2

3 3

1 1

3 32 2

3 52 ( 5 )

( 5 )

3 5 2 ( 5 ) 3 5 2 1 0

( 5 ) ( 5 )

5 1 5 5 ( 3 )

( 5 ) ( 5 )

x xx x

x x

x x x x x x x x

x x x x

x x x x

x x x x

1 / 2

1 1

2 2

11 / 2

22

1 11

2 2

1

21 1

2 2

2 1Q .6 ) ( )

( 2 3 )

( 2 3 ) ( 2 1) ( 2 1) ( 2 3 )2 1

( ) [ ]( 2 3 )

{ ( 2 3 ) }

1( 2 3 ) ( 2 ) ( 2 1) . .( 2 3 ) ( 2 3 )

2

( 2 3 )

( 2 1)2 ( 2 3 )1

2 ( 2 3 ) ( 2 1)( 2 3 ) ( 2 )(2

( 2 3 )

xf x

x

d dx x x x

d x d x d xf x

d x xx

dx x x x

d x

x

xx

x x x

x

1

2

1

2

1 33

2 2

2 3)

( 2 3 )

2 ( 2 3 ) ( 2 1)

4 6 2 1 2 7 2 7( 2 3 )

( 2 3 ) ( 2 3 )( 2 3 ) ( 2 3 ) ( 2 3 )

x

x

x x

x x x xx

x xx x x


0 .4

0 .4 0 .4

0 .4 0 .4 0 .4

Q .7 ) ( ) 2 .5 ; [ ( ) ]

( ) [ 2 .5 ] 2 .5 ( )

2 .5 ( 0 .4 ) 2 .5 ( 0 .4 )

x x x

x x

x x x

df x e e e

d x

d df x e e

d x d x

de x e e

d x

0 .3 6

0 .3 6 0 .3 6

0 .3 6 0 .3 6 0 .3 6

Q .8 ) ( ) 1 0

( ) [1 0 ] 1 0 ( )

1 0 (0 .3 6 ) 1 0 (0 .3 ) 3

x

x x

x x x

f x e

d df x e e

d x d x

de x e e

d x

0 .3

0 .3 0 .3

0 .3 0 .3

Q .9 ) ( ) (0 .4 ) ; [ ( ) ln ]

( ) [(0 .4 ) ] (0 .4 ) . ln (0 .4 ) ( 0 .3 )

(0 .4 ) . ln (0 .4 )( 0 .3 ) 0 .2 7 4 9 (0 .4 )

x x x

x x

x x

df x a a a

d x

d df x x

d x d x

Q .1 0 ) ( ) 5 0 0 (1 .0 8 ) ; [ ( ) ln ]

( ) [5 0 0 (1 .0 8 ) ] 5 0 0 [(1 .0 8 ) ] 5 0 0 .(1 .0 8 ) . ln (1 .0 8 ) ( )

5 0 0 .(1 .0 8 ) . ln (1 .0 8 )( 1) 3 8 .4 8 (1 .0 8 )

x x x

x x x

x x

df x a a a

d x

d d df x x

d x d x d x

1 / 3 1 1Q .1 1) ( ) ln ( 2 5 ) = ln ( 2 5 ); [ ( ln ) ]

3

1 1 1 ( ) [ ln ( 2 5 )] . ( 2 5 )

3 3 ( 2 5 )

1 1 2. ( 2 )

3 ( 2 5 ) 3 ( 2 5 )

df x x x x

d x x

d df x x x

d x x d x

x x

3

3 3

3

2

2 2

2 2 2

1Q .1 2 ) ( ) ln ( 2 6 ); [ ( ln ) ]

1 ( ) [ ln ( 2 6 )] ( 2 6 )

( 2 6 )

1 1 3( 1)(6 6 ) .6 ( 1)

2 ( 3 ) 2 ( 3 ) ( 3 )

df x x x x

d x x

d df x x x x x

d x x x d x

xx x

x x x x x x


1

21 1

Q .1 3) ( ) ln (3 2 ) = ln (3 2 ); [ ( ln ) ]2

1 1 ( ) [ ln (3 2 )] [ ln (3 2 )]

2 2

1 1 1 1 3. (3 2 ) . (3 )

2 (3 2 ) 2 (3 2 ) 2 (3 2 )

df x x x x

d x x

d df x x x

d x d x

dx

x d x x x

Partial Differentiation Problems

2 2

2 2

2 2

Q .1 4 ) ( , ) 3 2 4

[ ( , ) ] [3 2 4 ] 6 2

( ) ( 6 2 ) 6

[ ( , ) ] [3 2 4 ] 2 2

( ) ( 2 2 ) 2

F irs t w ith re s p e c t to x th e n w ith re s p e c t t

x

x x x

y

y y y

x y

f x y x x y y

f f x y x x y y x yx x

f f x yx x

f f x y x x y y x yy y

f f x yy y

f

2 2

o y

[ ( ) ] [ (3 2 4 )] (6 2 ) 2x y

f f x x y y x yy x y x y


1 / 2 1 / 3

11

1 / 2 1 / 3 1 / 2 1 / 23

2

3

21

1 / 2 3

2 5

3 3

1 / 2 1 / 3 1 / 3 1 / 3

1 1

2 2

Q .1 5 ) ( , ) 2 3

1[ ( , ) ] [ 2 3 ] 2 3 . 2

3

2 2( ) ( 2 ) .( )

33

1 1[ ( , ) ] [ 2 3 ] 2 . . 3 3

2

x

x x x

y

f x y x y x y

yf f x y x y x y y y x y

x xx

y yf f y y x

x xx x

xf f x y x y x y x x x

y yy y

f

1 / 3

1 3 3

2 2 2

1 / 2 1 / 3 1 / 2

2

3

1 2 1 2

2 3 2 3

1 1( ) ( 3 ) ( )

22

F irs t w ith re s p e c t to x th e n w ith re s p e c t to y

[ ( ) ] [ ( 2 3 )] ( 2 )

1 1 1 1 12 .( ) .

2

y y y

x y

x y

x xf x x

y yy y y

f

yf f x y x y y

y x y x yx

y yx x


Applied Mathematics


Problem Set 8-1:

For each of the following find the coordinates of all local and endpoint optimum (maximum or

minimum) points, and all stationary inflection points.

Question No. 1:

2

2 2

2

( ) 5

( ) [5 ] (5 ) ( ) 5 2

A t s ta t io n a ry p o in t ( ) 0

. .5 2 0

, 5 / 2

5 5 2 5 2 5 5 0 2 5 2 5If 5 / 2 th e n y= (5 / 2 ) 5 ( ) ( )

2 2 2 4 4 4

5 2 5T h e re is o n ly o n e s ta t io n a ry p o in t ( , ) ( 2 .5 , 6 .

2 4

f x x x

d d df x x x x x x

d x d x d x

f x

i e x

o r x

x f

o r

2 5 )

If 2 th e n ( 2 ) 5 2 ( 2 ) 1 , (+ v e )

T h e fu n c t io n is in c re a s in g b e fo re th e p o in t

If 3 th e n (3 ) 5 2 (3 ) 1 , ( v e )

T h e fu n c t io n is d e c re a s in g a f te r th e p o i n t

T h e re fo re w e c a n c o n c lu d e th a t th e s ta t i o n a

x f

x f

ry p o in t (2 .5 , 6 .2 5 ) is a

lo c a l m a x im u m a n d a s w e h a v e o n ly o n e s t a t io n a ry p o in t , th e re is n o

m in im a o r s ta t io n a ry in f le c t io n p o in t .

Question No. 2:

2

2 2

( ) 2 1 2 2 0

( ) [ 2 1 2 2 0 ] 2 ( ) 1 2 ( ) ( 2 0 ) 4 1 2


. .4 1 2 0

, 3

f x x x

d d d df x x x x x x

d x d x d x d x

f x

i e x

o r x


2If 3 th e n y= (3 ) 2 (3 ) 1 2 (3 ) 2 0 2

T h e re is o n ly o n e s ta t io n a ry p o in t (3 , 2 )

If 2 th e n ( 2 ) 4 ( 2 ) 1 2 4 , ( v e )

T h e fu n c tio n is d e c re a s in g b e fo re th e p o in t

If 4 th e n ( 4 ) 4 ( 4 ) 1 2 4 , ( v e )

T h e fu n c tio

x f

x f

x f

n is in c re a s in g a f te r th e p o in t

T h e re fo re w e c a n c o n c lu d e th a t th e s ta t i o n a ry p o in t (3 , 2 ) is a

lo c a l m in im u m a n d a s w e h a v e o n ly o n e s t a t io n a ry p o in t , th e re is n o

m a x im a o r s ta t io n a ry in f le c t io n p o in t .

Do the following at home: Q. No. 3: 2

( ) 3f x x Q. No. 4: 2

( ) 2 2 0f x x

Question No. 5:

2

2 2

2

( ) 3 0 3 1 0

( ) [3 0 3 1 0 ] (3 0 ) (3 ) (1 0 ) 3 0 6


. .3 0 6 0

, 5

If 5 th e n y= (5 ) 3 0 (5 ) 3 (5 ) 1 0 1 5 0 7 5 1 0 8 5

T h e re is o n ly o n e s ta t io n a ry p o in t (5 , 8 5 )

f x x x

d d d df x x x x x x

d x d x d x d x

f x

i e x

o r x

x f

If 4 th e n ( 4 ) 3 0 6 ( 4 ) 6 , ( v e )

T h e fu n c tio n is in c re a s in g b e fo re th e p o in t

If 6 th e n (6 ) 3 0 6 (6 ) 6 , ( v e )

T h e fu n c tio n is d e c re a s in g a f te r th e p o i n t

T h e re fo re w e c a n c o n c lu d e th a t th e s ta t i o n a r

x f

x f

y p o in t (5 , 8 5 ) is a

lo c a l m a x im u m a n d a s w e h a v e o n ly o n e s t a t io n a ry p o in t , th e re is n o

m in im a o r s ta t io n a ry in f le c t io n p o in t .


Do it at home: Q. No. 6: 2

( ) 1f x x x

Question No. 7:

3 2

3 2 3 2 2

2

3

( ) 1 2 1 2

( ) [ 1 2 1 2 ] ( ) 1 2 ( ) (1 2 ) 3 2 4


. .3 2 4 0

, 3 ( 8 ) 0

, ( 8 ) 0

E ith e r , 0 o r , ( 8 ) 0

. 0 , 8

If 0 th e n y= (0 ) (0 ) 1 2

f x x x

d d d df x x x x x x x

d x d x d x d x

f x

i e x x

o r x x

o r x x

x x

i e x

x f

2

3 2

2

( 0 ) 1 2 1 2

If 8 th e n y = (8 ) (8 ) 1 2 (8 ) 1 2 2 4 4

T h e re a re tw o s ta t io n a ry p o in ts (0 ,1 2 ) a n d (8 , 2 4 4 )

F irs t le t u s c h e c k s ta t io n a ry p o in t (0 ,1 2 )

If 1 th e n ( 1) 3 ( 1) 2 4 ( 1) 2 7 , ( v e )

T h e fu n c tio n is

x f

x f

2

in c re a s in g b e fo re th e p o in t

If 1 th e n (1) 3 (1) 2 4 (1) 2 1 , ( v e )


T h e re fo re w e c a n c o n c lu d e th a t th e s ta t i o n a ry p o in t (0 ,1 2 ) is a

lo c a l m a x im u m

N o w le t u s c

x f

2

2

h e c k s ta t io n a ry p o in t (8 , 2 4 4 )

If 7 th e n (7 ) 3 ( 7 ) 2 4 (7 ) 2 1 , ( v e )


If 9 th e n (9 ) 3 (9 ) 2 4 (9 ) 2 7 , ( v e )

T h e fu n c tio n is in c re a s in g a f te r th e p o i n t

T h e r

x f

x f

e fo re w e c a n c o n c lu d e th a t th e s ta t io n a r y p o in t (8 , 2 4 4 ) is a

lo c a l m in im u m a n d a s th e re is n o o th e r s ta t io n a ry p o in t ,

th e re fo re th e re is n o s ta t io n a ry in f le c t io n p o in t .

Do it at home: Q. No. 8: 3 2

( ) 3 2f x x x


Question No. 9:

3

3 3 2

2

3

2

( ) 3 2 7

( ) [3 2 7 ] 3 ( ) ( 2 7 ) 9


. .9 0

, 0

If 0 th e n y= (0 ) 3 (0 ) 2 7 2 7

T h e re is o n e s ta t io n a ry p o in t (0 , 2 7 )

If 1 th e n ( 1) 9 ( 1) = 9 , ( v e )

T h e fu

f x x

d d df x x x x

d x d x d x

f x

i e x

o r x

x f

x f

2

n c tio n is in c re a s in g b e fo re th e p o in t

If 1 th e n (1) 9 (1) = 9 , ( v e )


T h e re fo re w e c a n c o n c lu d e th a t th e s ta t i o n a ry p o in t (0 , 2 7 ) is a

s ta t io n a ry in f le c t io n p o

x f

in t a n d a s th e re is n o o th e r s ta t io n a ry p o in t ,

th e re fo re th e re is n o m in im a o r m a x im a .

Question No. 10:

3

3 3 2

2

3

2

( ) 1 4 2

( ) [1 4 2 ] (1 4 ) 2 ( ) 6


. . 6 0

, 0

If 0 th e n y= (0 ) 1 4 2 (0 ) 1 4

T h e re is o n e s ta t io n a ry p o in t (0 ,1 4 )

If 1 th e n ( 1) 6 ( 1) = 6 , ( v e )

T h

f x x

d d df x x x x

d x d x d x

f x

i e x

o r x

x f

x f

2

e fu n c tio n is d e c re a s in g b e fo re th e p o in t

If 1 th e n (1) 6 (1) = 6 , ( v e )


T h e re fo re w e c a n c o n c lu d e th a t th e s ta t i o n a ry p o in t (0 ,1 4 ) is a

s ta t io n a ry in f le c t i

x f

o n p o in t a n d a s th e re is n o o th e r s ta t io n a ry p o in t ,

th e re fo re th e re is n o m in im a o r m a x im a .


Do these at home: Q. No. 11: 2 3

( ) 2 1 2 6 0 2 0f x x x x

Q. No. 12: 2 3

( ) 6 9f x x x x Q. No. 13:

1

2( ) 2 8f x x x

Question No. 14:

2

3

2 2 2 11

3 3 3 3

1

3

1

3

1

3

1

3

1

3 33

( ) 9 2

2 ( ) [9 2 ] 9 ( ) 2 ( ) 9 . ( ) 2 6 2

3


. .6 2 0

6, 2

6,

2

, 3

, ( ) 3 ; [ ta k in g p o w e r 3 in b o th s id e ]

, 2 7

If 2 7 th e

f x x x


d x d x d x

f x

i e x

o r

x

o r x

o r x

o r x

o r x

x

2 2

33 3

2

1

3

n y= ( 2 7 ) 9 ( 2 7 ) 2 ( 2 7 ) 9 (3 ) 5 4

9 (3 ) 5 4 2 7

T h e re is o n e s ta t io n a ry p o in t (2 7 , 2 7 )

If 2 6 th e n ( 2 6 ) 6 ( 2 6 ) 2 0 .0 2 5 , ( v e )

T h e fu n c tio n is in c re a s in g b e fo re th e

f

x f

1

3

p o in t

If 2 8 th e n ( 2 8 ) 6 ( 2 8 ) 2 0 .0 2 4 , ( v e )


T h e re fo re w e c a n c o n c lu d e th a t th e s ta t i o n a ry p o in t (2 7 , 2 7 ) is a

lo c a l m a x im u m a n d a s th e re is n o o th e r s ta

x f

t io n a ry p o in t ,

th e re fo re th e re is n o m in im a a n d s ta t io n a ry in f le c t io n p o in t .


Do these at home: Q. No. 15:

3 1

2 2( ) 1 2f x x x

Q. No. 16:

1 2

3 3( ) 1 0f x x x Q. No. 17:

6 4( ) 4f x x

x

Q. No. 18: 2

2 5 0( ) 4f x x

x

Question No. 19:

3 2

3 2 3 2

2

( ) 9 2 7 ; [ -1 ,6 ]

T h e fu n c t io n is re s tr ic te d to a ll v a lu e s o f x

b e tw e e n -1 to 6 a n d in c lu s iv e .

( ) [ 9 2 7 ] ( ) 9 ( ) 2 7 ( )

3 1

f x x x x


d x d x d x d x

x

2

2

2 2

2

3 2

2

8 2 7


. .3 1 8 2 7 0

, 3 ( 6 9 ) 0

, 2 . .3 3 0

, ( 3 ) 0

, 3

If 3 th e n y= (3 ) 3 9 (3 ) 2 7 (3 ) 2 7

T h e re is o n e s ta t io n a ry p o in t (3 , 2 7 )

If 2 th e n ( 2 ) 3 ( 2 ) 1 8 ( 2 ) 2 7

x

f x

i e x x

o r x x

o r x x

o r x

o r x

x f

x f

2

3, ( v e )


If 4 th e n ( 4 ) 3 ( 4 ) 1 8 ( 4 ) 2 7 3 ( v e )

T h e fu n c t io n is in c re a s in g a f te r th e p o in t

T h e re fo re w e c a n c o n c lu d e th a t th e s ta t i o n a ry p o in t (3 , 2 7 ) is a

s t

x f

a t io n a ry in f le c t io n p o in t .


3 2

2

E n d p o in t A n a lys is :

T h e le f t h a n d e n d p o in t: 1,

y= ( 1) ( 1) 9 ( 1) 2 7 ( 1) 3 7

T h e le f t h a n d e n d p o in t is ( 1, 3 7 )

If 0 .5 th e n ( 0 .5 ) 3 ( 0 .5 ) 1 8 ( 0 .5 ) 2 7 3 5 .2 5 , ( v e )

T h e fu n c tio n is in c re a s in g a f

x

f

x f

3 2

te r th e p o in t

T h e re fo re w e c a n c o n c lu d e th a t th e le f t h a n d e n d p o in t ( 1, 3 7 )

is a n e n d p o in t m in im u m .

T h e r ig h t h a n d e n d p o in t: 6 ,

y= (6 ) (6 ) 9 (6 ) 2 7 (6 ) 5 4

T h e r ig h t h a n d e n d p o in t is ( 6 , 5 4 )

If 5 th e

x

f

x

2

n (5 ) 3 (5 ) 1 8 (5 ) 2 7 1 2 , ( v e )


T h e re fo re w e c a n c o n c lu d e th a t th e r ig h t h a n d e n d p o in t ( 6 , 5 4 )

is a n e n d p o in t m a x im u m .

f

Question No. 20:

3 2

3 2 3 2

( ) 1 5 7 5 ; [ -8 ,2 ]

T h e fu n c tio n is re s tr ic te d to a ll v a lu e s o f x

b e tw e e n -8 to 2 a n d in c lu s iv e .

( ) [ 1 5 7 5 ] ( ) 1 5 ( ) 7 5 ( )

3

f x x x x


d x d x d x d x

x

2

2

2

2 2

2

3 0 7 5


. .3 3 0 7 5 0

, 3 ( 1 0 2 5 ) 0

, 2 . .5 5 0

, ( 5 ) 0

, 5

x

f x

i e x x

o r x x

o r x x

o r x

o r x


3 2

2

2

If 5 th e n y= ( 5 ) ( 5 ) 1 5 ( 5 ) 7 5 ( 5 ) 1 2 5

T h e re is o n e s ta t io n a ry p o in t ( 5 , 1 2 5 )

If 6 th e n ( 6 ) 3 ( 6 ) 3 0 ( 6 ) 7 5 3, ( v e )


If 4 th e n ( 4 ) 3 ( 4 )

x f

x f

x f

3 0 ( 4 ) 7 5 3 ( v e )

T h e fu n c tio n is in c re a s in g a f te r th e p o in t

T h e re fo re w e c a n c o n c lu d e th a t th e s ta t io n a ry p o in t ( 5 , 1 2 5 ) is a

s ta t io n a ry in f le c t io n p o in t .

3 2

2


T h e le f t h a n d e n d p o in t : 8 ,

y= ( 8 ) ( 8 ) 1 5 ( 8 ) 7 5 ( 8 ) 1 5 2

T h e le f t h a n d e n d p o in t is ( 8 , 1 5 2 )

If 7 th e n ( 7 ) 3 ( 7 ) 3 0 ( 7 ) 7 5 1 2 , ( v e )

T h e fu n c tio n is in c re a s in g a f te r th e

x

f

x f

3 2

p o in t

T h e re fo re w e c a n c o n c lu d e th a t th e le f t h a n d e n d p o in t ( 8 , 1 5 2 )


T h e r ig h t h a n d e n d p o in t : 2 ,

y= ( 2 ) ( 2 ) 1 5 ( 2 ) 7 5 ( 2 ) 2 1 8

T h e r ig h t h a n d e n d p o in t is ( 2 , 2 1 8 )

If 1 th e n

x

f

x f

2

(1) 3 (1) 3 0 (1) 7 5 1 0 8 , ( v e )


T h e re fo re w e c a n c o n c lu d e th a t th e r ig h t h a n d e n d p o in t ( 2 , 2 1 8 )

is a n e n d p o in t m a x im u m .

Do these at home: Q. No. 21: 4

( ) 3 2 1 0 0; [ 2 , 4 ]f x x x

Q. No. 22: 4

( ) 1 0 8 ; [ 1 , 5 ]f x x x


Question No. 23:

3 4

3 4 3 4

( ) 4 3 4 ; [ 1 ,2 )

T h e fu n c tio n is re s tr ic te d to a ll v a lu e s o f x

b e tw e e n 1 to 2 , in c lu d in g 1 b u t e x c lu d in g 2 .

( ) [ 4 3 4 ] 4 ( ) 3 ( ) ( 4 )

f x x x

d d d df x x x x x

d x d x d x d x

2 3

2 3

2

2

1 2 1 2


. .1 2 1 2 0

,1 2 (1 ) 0

, (1 ) 0

, 0 ,1 0

, 1

x x

f x

i e x x

o r x x

o r x x

E ith e r x o r x

o r x

3 4

3 4

2 3

If 0 th e n y= (0 ) 4 (0 ) 3 (0 ) 4 4

If 1 th e n y= (1) 4 (1) 3 (1) 4 5

T h e re a re tw o s ta t io n a ry p o in ts ( 0 , 4 ) a n d (1, 5 )

F irs t le t u s c h e c k s ta t io n a ry p o in t ( 0 , 4 )

If 0 .5 th e n ( 0 .5 ) 1 2 ( 0 .5 ) 1 2 ( 0 .5 )

x f

x f

x f

2 3

4 .5 , ( v e )


If 0 .5 th e n (0 .5 ) 1 2 (0 .5 ) 1 2 (0 .5 ) 1 .5 , ( v e )

T h e fu n c t io n is in c re a s in g a f te r th e p o i n t

T h e re fo re w e c a n c o n c lu d e th a t th e s ta t i o n a ry p o in t ( 0 ,

x f

2 3

2

4 ) is a

s ta t io n a ry in f le c t io n p o in t .

N o w le t u s c h e c k s ta t io n a ry p o in t (1, 5 )

If 0 .5 th e n (0 .5 ) 1 2 (0 .5 ) 1 2 (0 .5 ) 1 .5 , ( v e )


If 1 .5 th e n (1 .5 ) 1 2 (1 .5 ) 1

x f

x f

3

2 (1 .5 ) 1 3 .5 , ( v e )

T h e fu n c t io n is d e c re a s in g a f te r th e p o i n t

T h e re fo re w e c a n c o n c lu d e th a t th e s ta t i o n a ry p o in t (1, 5 ) is a

lo c a l m a x im u m .


3 4


T h e re is n o r ig h t h a n d e n d p o in t , h e n c e n o r ig h t h a n d e n d p o in t o p tim a

T h e le f t h a n d e n d p o in t: 1,

y= ( 1) 4 ( 1) 3 ( 1) 4 3

T h e le f t h a n d e n d p o in t is ( 1, 3 )

If 0 .5 th e n ( 0 .5 )

x

f

x f

2 3

1 2 ( 0 .5 ) 1 2 ( 0 .5 ) 4 .5 , ( v e )


T h e re fo re w e c a n c o n c lu d e th a t th e le f t h a n d e n d p o in t ( 1, 3 )


Do these at home: Q. No. 24: 4 3

( ) 4 3 0 ; [ 2 , 4 )f x x x

Question No. 25: When x gallons of antifreeze are produced, the average cost per gallon is

A(x) where,1 0 0

( ) 0 .0 4 1A x xx

a) How many gallons should be produced if average cost per gallon is to be minimized?

1 0 0( ) 0 .0 4 1

W e n e e d to f in d th e m in im u m p o in t (s ta t i o n a ry p o in t)

to f in d o u t th e m in im u m a v e ra g e c o s t

1 0 0 1 ( ) [ 0 .0 4 1] 1 0 0 ( ) 0 .0 4 ( ) (1)

A x xx

d d d dA x x x

d x x d x x d x d x

2

2

2

2

2

1 1 0 0 0 .0 4


1. . 1 0 0 0 .0 4 0

1,1 0 0 0 .0 4

, 0 .0 4 1 0 0

1 0 0, 2 5 0 0

0 .0 4

, 5 0 g a llo n

x

A x

i ex

o rx

o r x

o r x

o r x

When 50 gallons of antifreeze are produced, the average cost per gallon will be minimized.


b) Prove (a) is minimum

2

2

W e fin d th e s ta t io n a ry p o in t w h e re 5 0

1If 4 9 th e n A ( 4 9 ) 1 0 0 0 .0 4 0 .0 0 1, ( v e )

( 4 9 )


1If 5 1 th e n A (5 1) 1 0 0 0 .0 4 0 .0 0 1, ( v e )

(5 1)

T h e fu n c tio n is in c r

x

x

x

e a s in g a f te r th e p o in t

T h e re fo re w e c a n c o n c lu d e th a t th e s ta t i o n a ry p o in t w h e re 5 0 g a llo n

m a k e s th e a v e ra g e c o s t m in im u m .

x

c) Compute the minimum average cost per gallon.

W h e n 5 0 g a llo n , th e m in im u m a v e ra g e c o s t p e r g a llo n

1 0 0(5 0 ) 0 .0 4 (5 0 ) 1 $ 5 p e r g a llo n

5 0

x

A

Question No. 26: If we use 100 linear feet of fence to enclose a rectangular plot of land, we

need 2x feet for one pair of sides ( or x feet for one of these sides), leaving (100-2x) feet for

the other pair of sides (or (50-x) feet for one of these sides. The rectangular is then x feet

by (50-x) feet and its area, A(x) is,2

( ) (5 0 ) 5 0A x x x x x

a) Find the value of x that maximizes the area of the plot?

2

2 2

( ) 5 0

W e n e e d to f in d th e m a x im u m p o in t (s ta t i o n a ry p o in t)

to f in d o u t th e m a x im u m a re a

( ) [5 0 ] 5 0 ( ) ( )

5 0 2

A t s ta t io n a ry p o in t ( )

A x x x

d d dA x x x x x

d x d x d x

x

A x

0

. .5 0 2 0

, 2 5 fe e t

i e x

o r x

x x

50-x

50-x


b) Prove (a) is a maximum.

W e fin d th e s ta t io n a ry p o in t w h e re 2 5

If 2 4 th e n A ( 2 4 ) 5 0 2 ( 2 4 ) 2 , ( v e )


If 2 6 th e n A ( 2 6 ) 5 0 2 ( 2 6 ) 2 , ( v e )


x

x

x

T h e re fo re w e c a n c o n c lu d e th a t th e s ta t i o n a ry p o in t w h e re 2 5 fe e t

m a k e s th e a re a m a x im u m .

x

c) What are the dimensions of the maximum area rectangle?

Length = x feet = 25 feet

Width = 50—x = 50—25 = 25 feet

i.e it’s a square

d) What is the maximum area?

W h e n 2 5 fe e t , th e a re a w ill b e

( 2 5 ) 2 5 (5 0 2 5 ) 6 2 5 s q u a re -fe e t

x

A

Question No. 27: In many sample surveys only a small number n of a large population of

people are interviewed. Published state results state what proportion, p, of the n people

interviewed answered yes to a particular question. Inherently, because only a fraction of

the population is interviewed, the proportion p is expected to be in error by V(p), where

V stands for variance, and

2

( )p p

V pn n

a) Find the value of p that maximizes V(p). 2

2

2

( )


to f in d o u t th e m a x im u m e rro r

1 1 1 2 ( ) [ ] ( ) ( )


1 2 2. . 0 ,

p pV p

n n

d p p d d pV p p p

d p n n n d p n d p n n

V p

pi e o r

n n

1

, 2 1 , 1 / 2

p

n n

o r p o r p



W e fin d th e s ta t io n a ry p o in t w h e re 1 / 2 0 . 5

1 2 (0 .3 ) 1 0 .6 0 .4If 0 .3 th e n V (0 .3 ) = , ( v e )


1 2 (0 .6 ) 1 1 .2 0 .2If 0 .6 th e n V (0 .6 ) = , ( v e )

T h e fu n c t

p

pn n n n

pn n n n

io n is d e c re a s in g a f te r th e p o in t

T h e re fo re w e c a n c o n c lu d e th a t th e s ta t i o n a ry p o in t w h e re 0 .5

m a k e s th e v a r ia n c e m a x im u m .

p

c) Taking n=10, find V(0.5) and V(0.1) 2

2

0 .5 0 .5(0 .5 ) 0 .0 2 5

1 0 1 0

0 .1 0 .1(0 .1) 0 .0 0 9

1 0 1 0

V

V

Question No. 28: The profit realized when y gallons of distilled water are made and sold is

2( ) 2 0 0 .0 0 5P y y y

a) Find the number of gallons that should be made to maximize profit.

2

2 2

( ) 2 0 0 .0 0 5


to f in d o u t th e m a x im u m p ro f i t

( ) [ 2 0 0 .0 0 5 ] 2 0 ( ) 0 .0 0 5 ( )

2 0 0 .0 1

A t s ta t

P y y y

d d dP y y y y y

d y d y d y

y

io n a ry p o in t ( ) 0

. .2 0 0 .0 1 0 , 0 .0 1 2 0

2 0, = 2 0 0 0 g a llo n s

0 .0 1

P y

i e y o r y

o r y



W e fin d th e s ta t io n a ry p o in t w h e re 2 0 0 0

If 1 9 9 9 th e n P (1 9 9 9 ) 2 0 0 .0 1(1 9 9 9 ) 0 .0 1, ( v e )


If 2 0 0 1 th e n P ( 2 0 0 1) 2 0 0 .0 1( 2 0 0 1) 0 .0 1, ( v e )

T h e fu n c tio n is

y

y

p

d e c re a s in g a f te r th e p o in t

T h e re fo re w e c a n c o n c lu d e th a t th e s ta t i o n a ry p o in t w h e re

2 0 0 0 g a llo n s m a k e s th e p ro f i t m a x im u m .y

c) Compute the maximum profit

2( 2 0 0 0 ) 2 0 ( 2 0 0 0 ) 0 .0 0 5 ( 2 0 0 0 ) $ 2 0 , 0 0 0P


Applied Mathematics


Problem Set 8-2:

Find the first three derivative of the function

Question No. 1:

5 4 3

5 4 3 5 1 4 1 3 1 4 3 2

4 3 2 3 2

3 2 2

( ) 2 3

( ) [ 2 3 ] 5 2 .4 3 5 8 3

( ) [ ( )] (5 8 3 ) 2 0 2 4 6

( ) [ ( )] ( 2 0 2 4 6 ) 6 0 4 8 6

f x x x x

df x x x x x x x x x x

d x

d df x f x x x x x x x

d x d x

d df x f x x x x x x

d x d x

Do these at home: Q.2: 3 2

( ) 1f x x x x Q.3: 3 2

( ) 8 2f x x x

Q.4: 6 4 2

( ) 3 7f x x x x Q.5: 4 3 2

( ) 3 5 2f x x x x

Q.6: 2 2

( ) 5f x x x x x Q.7: 3

( ) ( 5 )f x x Q.8:

1

2 2( ) ( 1)f x x

Q.9:

3 4( )

3

xf x

x

Q.10:

2( )

2

xf x

x

Question No. 11:

2

2

( ) 4 3

( ) [ 4 3 ] 2 4 ; ( ) [ 2 4 ] 2 ;

( ) [ 2 ] 0

O p tim u m A n a lys is :


. .2 4 0

, 2

f x x x

d df x x x x f x x

d x d x

df x

d x

f x

i e x

o r x


2If 2 th e n y= ( 2 ) ( 2 ) 4 ( 2 ) 3 1

T h e re is o n ly o n e s ta t io n a ry p o in t ( 2 , 1)

( 2 ) 2 ; [+ v e , i .e . c o n c a v e u p

T h e re fo re w e c a n c o n c lu d e th a t th e s ta t io n a ry p o in t ( 2 , 1) is a

lo c a l m in im u m .

In f le c t io n P o in t

x f

f

A n a lys is :

( ) 0

, 2 0

. . n o in f le c t io n p o in t

f x

o r

i e

Question No. 12:

2

2

( ) 1 0 9

( ) [ 1 0 9 ] 2 1 0 ; ( ) [ 2 1 0 ] 2 ;

( ) [ 2 ] 0



. .2 1 0 0

, 5

f x x x

d df x x x x f x x

d x d x

df x

d x

f x

i e x

o r x

2If 5 th e n y= ( 5 ) ( 5 ) 1 0 ( 5 ) 9 3 4

T h e re is o n ly o n e s ta t io n a ry p o in t ( 5 , 3 4 )

( 5 ) 2 ; [+ v e , i .e . c o n c a v e u p

T h e re fo re w e c a n c o n c lu d e th a t th e s ta t io n a ry p o in t ( 5 , 3 4 ) is a


In f le

x f

f

c t io n P o in t A n a lys is :

( ) 0

, 2 0

. . n o in f le c t io n p o in t

f x

o r

i e


Question No. 13:

3 2

3 2 2

2

2

2

2

( ) 6 9 1

( ) [ 6 9 1] 3 1 2 9

( ) [3 1 2 9 ] 6 1 2 ;

( ) [6 1 2 ] 6



. .3 1 2 9 0

, 3 ( 4 3 ) 0

, 3 3 0

, ( 3 )(

f x x x x

df x x x x x x

d x

df x x x x

d x

df x x

d x

f x

i e x x

o r x x

o r x x x

o r x x

1) 0

, 3,1o r x

3 2

3 2

If 3 th e n y = (3 ) (3 ) 6 (3 ) 9 (3 ) 1 1

If 1 th e n y= (1) (1) 6 (1) 9 (1) 1 5

T h e re a re tw o s ta t io n a ry p o in ts (3 ,1) a n d (1, 5 )

L e t u s c h e c k p o in t (3 ,1) f i rs t ,

(3 ) 6 (3 ) 1 2 6 ; [+ v e , i .e . c o n c a v e u p ]

T h e re

x f

x f

f

fo re w e c a n c o n c lu d e th a t th e s ta t io n a ry p o in t (3 ,1) is a


L e t u s c h e c k p o in t (1, 5 ) n o w ,

(1) 6 (1) 1 2 6 ; [ v e , i .e . c o n c a v e d o w n ]

T h e re fo re w e c a n c o n c lu d e th a t th e s ta t i o n a ry p o in t (1,

f

3 2

5 ) is a


In f le c t io n P o in t A n a ly s is :

( ) 0

, 6 1 2 0

, 2

T h e n , ( 2 ) 6 0 ; w e h a v e a n in f le c t io n p o in t a t 2

If 2 th e n y= ( 2 ) ( 2 ) 6 ( 2 ) 9 ( 2 ) 1 3

i .e . w e h a v e in f le c t io n p o in t a t ( 2 , 3

f x

o r x

o r x

f x

x f

)


Question No. 14:

3

3 2

2

2

2

2

2

( ) 9

( ) [ 9 ] 3 9

( ) [3 9 ] 6 ;

( ) [ 6 ] 6

O p tim u m A n a ly s is :


. .3 9 0

, 3 ( 3 ) 0

, 3 0

, 3

, 3 im a g in a ry o r u n re a l n o .

f x x x

df x x x x

d x

df x x x

d x

df x x

d x

f x

i e x

o r x

o r x

o r x

o r x

3

T h e re is n o s ta t io n a ry p o in t a n d h e n c e n o o p tim a .

In f le c t io n P o in t A n a lys is :

( ) 0

, 6 0

, 0

T h e n , (0 ) 6 0 ; w e h a v e a n in f le c t io n p o in t a t 0

If 0 th e n y= ( 2 ) (0 ) 9 (0 ) 0

i .e . w e h a v e in f le c t i

f x

o r x

o r x

f x

x f

o n p o in t a t (0 , 0 )


Question No. 15:

3 2

3 2 2

2

2

2

( ) 2 4 3

( ) [ 2 4 3 ] 3 4 4

( ) [3 4 4 ] 6 4 ;

( ) [6 4 ] 6

O p tim u m A n a ly s is :


. .3 4 4 0

, 3 6 2 4 0

, ( 2 )(3 2 ) 0

2, 2 ,

3

f x x x x

df x x x x x x

d x

df x x x x

d x

df x x

d x

f x

i e x x

o r x x x

o r x x

o r x

3 2

3 2

If 2 th e n y= ( 2 ) ( 2 ) 2 ( 2 ) 4 ( 2 ) 3 5

2 2 2 2 2If th e n y= ( ) ( ) 2 ( ) 4 ( ) 3 4 .4 8

3 3 3 3 3

T h e re a re tw o s ta t io n a ry p o in ts ( 2 , 5 ) a n d ( 0 .6 7 , 4 .4 8 )

L e t u s c h e c k p o in t ( 2 , 5 ) f irs t ,

( 2 ) 6 ( 2 ) 4 8; [+ v e ,

x f

x f

f

i .e . c o n c a v e u p ]

T h e re fo re w e c a n c o n c lu d e th a t th e s ta t i o n a ry p o in t ( 2 , 5 ) is a


L e t u s c h e c k p o in t ( 0 .6 7 , 4 .4 8 ) n o w ,

( 0 .6 7 ) 6 ( 0 .6 7 ) 4 8 .0 2 ; [ v e , i .e . c o n c a v e d o w n ]

T h e re fo re w e c a

f

n c o n c lu d e th a t th e s ta t io n a ry p o in t ( 0 . 6 7 , 4 .4 8 ) is a


In f le c t io n P o in t A n a lys is :

( ) 0

, 6 4 0

, 0 .6 7

T h e n , ( 0 .6 7 ) 6 0 ; w e h a v e a n in f le c t io n p o in t a t 0 .6 7

If 0 .6 7 th e n y= (0

f x

o r x

o r x

f x

x f

3 2

.6 7 ) (0 .6 7 ) 2 (0 .6 7 ) 4 (0 .6 7 ) 3 0 .2 7 7

i.e . w e h a v e in f le c t io n p o in t a t ( 0 .6 7 , 0 .2 7 7 )

Do these at home: Q.16 3 2

( ) 3 9 3f x x x x Q.17 3 2

( ) 1 2 4 5 2f x x x x


Do these at home: Q.18 4 3 2

( ) 8 2 4f x x x x

Q.19 4

( ) 0 .1( 1 0 ) 2 5 .6 3 4 0 .8f x x x

Question No. 20: 4 3 2

4 3 2 3 2

3 2 2

2

3 2

( ) 8 1 8 2 7

( ) [ 8 1 8 2 7 ] 4 2 4 3 6

( ) [ 4 2 4 3 6 ] 1 2 4 8 3 6 ;

( ) [1 2 4 8 3 6 ] 2 4 4 8



. .4 2 4 3 6 0

, 4 (

f x x x x

df x x x x x x x

d x

df x x x x x x

d x

df x x x x

d x

f x

i e x x x

o r x x

2

2 2

2

6 9 ) 0

, ( 2 . .3 3 ) 0

, ( 3 ) 0

, 0 , 3

x

o r x x x

o r x x

o r x

4 3 2

4 3 2

2

If 0 th e n y= (0 ) (0 ) 8 (0 ) 1 8 (0 ) 2 7 2 7

If 3 th e n y= (3 ) (3 ) 8 (3 ) 1 8 (3 ) 2 7 0

T h e re a re tw o s ta t io n a ry p o in ts ( 0 , 2 7 ) a n d (3, 0 )

L e t u s c h e c k p o in t ( 0 , 2 7 ) f irs t ,

( 0 ) 1 2 (0 ) 4 8 (0 ) 3 6 3 6 ; [+ v e , i

x f

x f

f

2

3 2

.e . c o n c a v e u p ]

T h e re fo re w e c a n c o n c lu d e th a t th e s ta t i o n a ry p o in t ( 0 , 2 7 ) is a


L e t u s c h e c k p o in t (3 , 0 ) n o w ,

(3 ) 1 2 (3 ) 4 8 (3 ) 3 6 0 ; [ te s t fa i le d ]

W h e n 2 th e n ( 2 ) 4 ( 2 ) 2 4 ( 2 ) 3 6 (

f

x f

3 2

2 ) 8; [+ v e ]


W h e n 4 th e n ( 4 ) 4 ( 4 ) 2 4 ( 4 ) 3 6 ( 4 ) 1 6 ; [+ v e ]


T h e re fo re w e c a n c o n c lu d e th a t th e p o in t (3 , 0 ) is a

s ta t

x f

io n a ry in f le c t io n p o in t .


2

2

2

In f le c t io n P o in t A n a ly s is :

( ) 0

,1 2 4 8 3 6 0

,1 2 ( 4 3 ) 0

, 3 3 0

, ( 3 )( 1) 0

, 3 ,1

T h e n , (3 ) 2 4 (3 ) 4 8 2 4 0 ; w e h a v e a n in f le c t i o n p o in t a t 3

(1) 2 4 (1) 4 8 2 4 0 ; w e h a v e a n

f x

o r x x

o r x x

o r x x x

o r x x

o r x

f x

f

4 3 2

in f le c t io n p o in t a t 1

If 1 th e n y= (1) (1) 8 (1) 1 8 (1) 2 7 1 6

i.e . w e h a v e s ta t io n a ry in f le c t io n p o in t a t (3 , 0 ) a n d

n o n -s ta t io n a ry in f le c t io n p o in t a t (1, 1 6 )

x

x f


Applied Mathematics

Applications of Differential Calculus

Question 1: A rectangular plot of land is to be enclosed by a fence. Fence for the east-west

sides cost $10 per running foot, while that for the north-south sides cost $5 per running

foot. What is the maximum area that can be enclosed if $1,500 is available for purchasing

the fence?

Answer: Let, length of East-West sides are x foot each and that of North-South sides are y foot

each. To fence the perimeter (2x+2y) of the plot we have $1,500.

2

2

2

2 ($ 1 0 ) 2 ($ 5 ) $ 1 5 0 0

, 2 0 1 0 1 5 0 0

1 5 0 0 2 0,

1 0

1 5 0 0 2 0 1 5 0 0 2 0A re a o f th e re c ta n g u la r p lo t A ( )= .

1 0 1 0

F o r th e a re a to b e m a x im u m ,

A ( )= 0

1 5 0 0 2 0, ( ) 0

1 0

1, . (1 5 0 0 2 0 ) 01 0

1,1 0

x y

o r x y

xo r y

x x xx x y x

x

d x xo r

d x

do r x x

d x

o r

(1 5 0 0 4 0 ) 0

,1 5 0 0 4 0 0

1 5 0 0, 3 7 .5 fo o t

4 0

1If 3 7 fo o t , th e n A (3 7 ) {1 5 0 0 4 0 (3 7 )} 2 0 ; (+ v e )

1 0

1If 3 8 fo o t , th e n A (3 8 ) {1 5 0 0 4 0 (3 8 )} 2 0 ; ( v e )

1 0

i .e . 3 7 .5 fo o t is m a k in g th e a re a m a x im u m

T h e re fo r

x

o r x

o r x

x

x

x

1 5 0 0 2 0e th e m a x im u m a re a x y .

1 0

1 5 0 0 2 0 (3 7 .5 )(3 7 .5 ) 2 8 1 2 .5 s q .fo o t

1 0

xx

North (y)

South (y)

East

(x)

West

(x)


Question 2: A rectangular manufacturing plant with a floor area of 16,875 sq-feet is to be

built on a straight road. The front of the plant must be set back 60 feet from the road and

buffer strips (grass and trees) 30 feet on each side and 20 feet at the back must be

provided as shown in the figure. Because of the high cost of land, builders seek to

minimize the total area (plant plus buffers).

Answer:

A re a o f th e m a n u fa c tu r in g p la n t 1 6 , 8 7 5 s q fe e t

1 6 8 7 5,

A re a o f th e to ta l p lo t A ( ) x 8 0 y 6 0 = 8 0 6 0 4 8 0 0

1 6 8 7 5 1 3 5 0 0 0 01 6 8 7 5 8 0 . 6 0 4 8 0 0 2 1 6 7 5 6 0

F o r th e a re a to b e m in im u m ,

A ( )= 0

,

x y

o r yx

x x y y x

x xx x

x

do r

2

2

2

2

2 3

1 3 5 0 0 0 0( 2 1 6 7 5 6 0 ) 0

1 3 5 0 0 0 0, 6 0 0

1 3 5 0 0 0 0, 6 0

1 3 5 0 0 0 0,

6 0

, 2 2 5 0 0

, 1 5 0 fo o t

If A (1 5 0 )> 0 , th e n th e v a lu e o f w ill m a k e th e a re a m in im u m

1 3 5 0 0 0 0 1 2A ( ) ( 6 0 ) 1 3 5 0 0 0 0 ( 2 )

xd x x

o rx

o rx

o r x

o r x

o r x

x

dx

d x x x

3

3

7 0 0 0 0 0

2 7 0 0 0 0 0A (1 5 0 ) 0 .8; [+ v e , c o n c a v e u p ]

(1 5 0 )

T h e re fo re th e m in im u m to ta l la n d a re a

1 3 5 0 0 0 0 2 1 6 7 5 6 0 (1 5 0 ) 3 9 , 6 7 5 s q -f t

1 5 0

x

30ft

30ft

x

60ft

20

ft

x

y

y

(x+80)ft

y

+

60

ft


Question 3: In normal operations, a plant employs 100 workers working eight hours a

day for a total of 100(8)= 800 labour hours of work per day. In normal operations,

productivity averages 30 units per labour-hour worked. Thus in a normal day,

production is,

P= (Labour hours worked)(Output per labour hour)=(800)(30)=24,000 units

If the work level (labour hours) is raised over 800, management estimates that average

output per labour-hour falls off at the rate of 2.5 units for each for each extra 100 labour

hours or by 0.025 units for each labour hour in excess of 800. What will be the

appropriate labour hour (x) that would maximize output?

Ans:

2

2

T o ta l o u tp u t w h e n x 8 0 0 ;

( ) [3 0 ( 8 0 0 )0 .0 2 5 ] (3 0 0 .0 2 5 2 0 ) 5 0 0 .0 2 5

F o r th e o u tp u t to b e m a x im u m ,

( ) = 0

, (5 0 0 .0 2 5 ) 0

, 5 0 0 .0 5 0

, 0 .0 5 5 0

5 0, 1 0 0 0 u n its

0 .0 5

( ) (5 0 0 .0 5 )

P x x x x x x x

P x

do r x x

d x

o r x

o r x

o r x

dP x x

d x

2

0 .0 5

(1 0 0 0 ) 0 .0 5; [ v e , c o n c a v e d o w n ]

A s P (1 0 0 0 )< 0 , th e n th e v a lu e o f w ill m a k e th e o u tp u t m a x im u m

T h e re fo re th e m a x im u m o u tp u t w ill b e

(1 0 0 0 ) 5 0 (1 0 0 0 ) 0 .0 2 5 (1 0 0 0 ) 2 5 , 0 0 0 u n its

P

x

P

Question 4: A consulting firm conducts training sessions for employees of various

companies. The charge to a company sending employees to a session is $50 per employee,

less $0.50 for each employee in excess of 10. The consulting firm further has a fixed total

charge for groups of x or more, where x is the number that maximizes the prorated group

charge. What should x be, and what is the maximum total group charge to a company?


2

2

N o . o f e m p lo ye e s in a g ro u p th a t m a x im iz e s th e g ro u p c h a rg e

T o ta l g ro u p c h a rg e w h e n 1 0 ;

( ) [5 0 ( 1 0 )0 .5 0 ] (5 0 0 .5 0 5 ) 5 5 0 .5 0

F o r th e g ro u p c h a rg e to b e m a x im u m ,

( ) = 0

, (5 5 0 .5 0 ) 0

x

x

C x x x x x x x

C x

do r x x

d x

o

, 5 5 0

, 5 5 e m p lo ye e s

( ) (5 5 ) 1

(5 5 ) 1; [ v e , c o n c a v e d o w n ]

A s (5 5 ) < 0 , th e n th e v a lu e o f w ill m a k e th e g ro u p c h a rg e m a x im u m

T h e re fo re th e m a x im u m to ta l g ro u p c h a rg e w ill b e

(5 5 )

r x

o r x

dC x x

d x

C

C x

C

2

5 5 (5 5 ) 0 .5 0 (5 5 ) $ 1 5 1 2 .5 fo r 5 5 e m p lo ye e s o r m o re

Question 5: When State College charges $195 for continuing education class in the uses of

microcomputers, it attracts 125 students. For each $10 decrease in the charge, an

additional 8 students will attend the class. Find the tuition value State College should

charge to maximize revenue and then find this maximum revenue.

2

2

T u it io n v a lu e c h a rg e d b y S ta te C o lle g e ; w h e n 1 9 5

1 9 5 1 9 5( ) [1 2 5 (8 )] [1 2 5 ( 4 )]

1 0 5

6 2 5 7 8 0 4 1 4 0 5 4( )

5 5

F o r th e re v e n u e to b e m a x im u m ,

( ) = 0

1 4 0 5 4, ( ) 0

5

1, (1 4 0 5 8 ) 0

5

,1 4 0 5 8

x x

x xR x x x

x x xx

R x

d x xo r

d x

o r x

o r

0

1 4 0 5, $ 1 7 5 .6 2 5

8

x

o r x


1 1 8( ) { (1 4 0 5 8 )} (1 4 0 5 8 )

5 5 5

8(1 7 5 .6 2 5 ) ; [ v e , c o n c a v e d o w n ]

5

A s (1 7 5 .6 2 5 ) < 0 , th e n th e v a lu e o f tu i t io n fe e $ 1 7 5 .6 2 5

w ill m a k e th e re v e n u e m a x im u m

T h e re fo re th e m a x im u m re v e n u e

d dR x x x

d x d x

R

R x

2

w ill b e

1 4 0 5 (1 7 5 .6 2 5 ) 4 (1 7 5 .6 2 5 )(1 7 5 .6 2 5 ) $ 2 4 , 6 7 5 .3 1

5R

Question 6: When x gallons of alcohol are produced, the average cost per gallon is A(x)

dollars, where,

2 0 0( ) 0 .0 5 ; 0

0 .1 5A x x x

x

a) Find the value of x where A(x) has a stationary point.

1

1 1

2

2

2

A t s ta t io n a ry p o in t ,

( ) 0

2 0 0, ( 0 .0 5 ) 0

0 .1 5

, 2 0 0 (0 .1 5 ) (0 .0 5 ) 0

, 2 0 0 ( 1)(0 .1 5 ) (0 .1 5 ) 0 .0 5 0

, 2 0 0 (0 .1 5 ) (0 .1) 0 .0 5 0

2 0, 0 .0 5

(0 .1 5 )

2 0, (0 .1 5 )

0 .0 5

, (0 .

A x

do r x

d x x

d do r x x

d x d x

do r x x

d x

o r x

o rx

o r x

o r

21 5 ) 4 0 0

, (0 .1 5 ) 2 0

E ith e r , o r ,

( 0 .1 5 ) 2 0 ( 0 .1 5 ) 2 0

1 5 2 5, 1 5 0 , 2 5 0

0 .1 0 .1

g a llo n s c a n n o t b e n e g a t iv e , h e n c e x = 1 5

x

o r x

x x

o r x o r x

x

0 g a llo n s


b)

2 2

2 1

3 3

3

( ) { 2 0 (0 .1 5 ) 0 .0 5} { 2 0 (0 .1 5 ) } (0 .0 5 )

2 0 ( 2 )(0 .1 5 ) (0 .1 5 ) 0

1 44 0 (0 .1)

(0 .1 5 ) (0 .1 5 )

4(1 5 0 ) 0 .0 0 0 5; [+ v e , c o n c a v e u p ]

{0 .1(1 5 0 ) 5 )

T h e v a lu e o f 1 5 0 g a llo n s o c c

d d dA x x x

d x d x d x

dx x

d x

x x

A

x

u rs a t lo c a l m in im u m o f A ( )x

c)

2 0 0(1 5 0 ) 0 .0 5 (1 5 0 ) $ 1 7 .5 p e r g a llo n

0 .1(1 5 0 ) 5A

Question 7: Based on historical data, a bus company uses the function

1

2( ) (3 0 .6 ) 0 .1 ; 0P x x x x to estimate the net weekly profit, in hundreds of

dollars, if a particular bus route is x miles long. How long should the route be to

maximize the net profit, and what is the maximum net profit?

1

2

1

2

11

2

1

2

1

2

1

2

F o r th e p ro f i t to b e m a x im u m ,

( ) = 0

, { (3 0 .6 ) 0 .1 } 0

, { (3 0 .6 ) } ( 0 .1 ) 0

1, (3 0 .6 ) (3 0 .6 ) 0 .1 0

2

1, (3 0 .6 ) ( 0 .6 ) 0 .1 0

2

0 .3, 0 .1

(3 0 .6 )

0 .3 0, (3 0 .6 )

0 .1

, (3

P x

do r x x

d x

d do r x x

d x d x

do r x x

d x

o r x

o r

x

o r x

o r

1

20 .6 ) 3x


1

2 22

1 11

2 2

3

2

3

2

3

2

, { (3 0 .6 ) } (3 )

, 3 0 .6 9

9 3, 1 0 m ile s

0 .6

1( ) {0 .3 (3 0 .6 ) 0 .1} 0 .3 ( )(3 0 .6 ) (3 0 .6 )

2

0 .0 90 .1 5 (3 0 .6 ) (0 .6 )

(3 0 .6 )

0 .0 9(1 0 ) ; [ v e , c o n c a v e d o w n

{3 0 .6 (1 0 )}

o r x

o r x

o r x

d dP x x x x

d x d x

x

x

P

1

2

]

T h e re fo re 1 0 m ile s m a k e s th e p ro f i t m a x i m u m a n d

th e m a x im u m n e t p ro f i t , (1 0 ) {3 0 .6 (1 0 )} 0 .1(1 0 )

2 (1 0 0 ) $ 2 0 0 p e r w e e k

x

P

Question 8: When y gallons of crude oil are produced the average cost per barrel is A(y),

where,

2 5 0 0( ) + 0 .1 6 y

0 .0 4 9A y

y

a) Find the value of y that minimizes average cost per barrel.

1

1 1

2

2

F o r th e a v e ra g e c o s t to b e m in im u m ,

( ) = 0

2 5 0 0, { + 0 .1 6 y} 0

0 .0 4 9

, { 2 5 0 0 ( 0 .0 4 9 ) } (0 .1 6 ) 0

, 2 5 0 0 ( 1)(0 .0 4 9 ) ( 0 .0 4 9 ) 0 .1 6 0

, 2 5 0 0 (0 .0 4 9 ) ( 0 .0 4 ) 0 .1 6 0

, 2 5 0 0 (0 .0 4 9 )

A y

do r

d y y

d do r y y

d y d y

do r y y

d y

o r y

o r y

2

2

2

( 0 .0 4 ) 0 .1 6 0

1 0 0, 0 .1 6

(0 .0 4 9 )

1 0 0, (0 .0 4 9 )

0 .1 6

, (0 .0 4 9 ) 6 2 5

, (0 .0 4 9 ) 2 5

o ry

o r y

o r y

o r y


E ith e r , o r ,

( 0 .0 4 9 ) 2 5 ( 0 .0 4 9 ) 2 5

2 5 9 2 5 9, ,

0 .0 4 0 .0 4

, 4 0 0 , 8 5 0

w e d is c a rd 8 5 0 b e c a u s e 0

( ) { 1 0 0 (0

y y

o r y o r y

o r y o r y

y y

dA y

d y

2 2 1

3

3

.0 4 9 ) 0 .1 6} 1 0 0 ( 2 )(0 .0 4 9 ) (0 .0 4 )

8

(0 .0 4 9 )

8( 4 0 0 ) 0 .0 0 0 5 1 2 ; [+ v e , c o n c a v e u p ]

{0 .0 4 ( 4 0 0 ) 9 )

4 0 0 g a llo n s w ill m a k e th e a v e ra g e c o s t p e r b a rre l m in im u m

y y

y

A

y

b) Compute the minimum average cost per barrel.

M in im u m a v e ra g e c o s t p e r b a rre l,

2 5 0 0( 4 0 0 ) + 0 .1 6 (4 0 0 ) $ 1 6 4 p e r b a rre l

0 .0 4 ( 4 0 0 ) 9A

Do this at home: Question 9: When x gallons of olive oil are produced the average cost

per barrel is A(x), where,

4 , 0 0 0( ) + 0 .2 5 x ; x > 0

0 .1 2 0A x

x

a) Find the value of x that minimizes average cost per barrel.

b) Compute the minimum average cost per barrel.

Question 10: Profit realized when x thousand gallons of antifreeze are produced and sold

is P(x) thousand dollars, where,

1

2( ) (1 0 0 1 0 ) 0 .2P x x x

a) Find the value of x that leads to maximum profit.

1

2

1

2

11

2

F o r th e p ro f i t to b e m a x im u m ,

( ) = 0

, { (1 0 0 1 0 ) 0 .2 } 0

, (1 0 0 1 0 ) (0 .2 ) 0

1, (1 0 0 1 0 ) (1 0 0 1 0 ) (0 .2 ) 0

2

P x

do r x x

d x

d do r x x

d x d x

d do r x x x

d x d x


1

2

1

2

1

2

1

2

1 11

2 2

1, (1 0 0 1 0 ) (1 0 ) 0 .2 0

2

5, 0 .2

(1 0 0 1 0 )

5, (1 0 0 1 0 )

0 .2

, (1 0 0 1 0 ) 2 5

, (1 0 0 1 0 ) 6 2 5

6 2 5 1 0 0, 5 2 .5 th o u s a n d g a llo n s

1 0

1( ) {5 (1 0 0 1 0 ) 0 .2} 5 ( )(1 0 0 1 0 ) (1 0 ) 2 5 (1 0 0 1 0

2

o r x

o r

x

o r x

o r x

o r x

o r x

dP x x x x

d x

3

2

3

2

)

(5 2 .5 ) 2 5{1 0 0 1 0 (5 2 .5 )} 0 .0 0 1 6 ; [ v e , c o n c a v e d o w n ]

T h u s 5 2 .5 th o u s a n d g a llo n s le a d s to m a x i m u m p ro f i t

P

x

b) Compute the maximum profit.

1

2(5 2 .5 ) {1 0 0 1 0 (5 2 .5 )} 0 .2 (5 2 .5 ) 1 4 .5 th o u s a n d d o lla rs $ 1 4 , 5 0 0P

Question 11: A company operates a fleet of delivery trucks. Study shows that gallons of

fuel consumed per mile of driving, F(x), is related to the speed at which a truck is driven,

x miles per hour, by the function

1

2

1 2

( ) ; 1 0 8 0

w h e re a n d a re p a ra m e te rs th a t v a ry s o m e w h a t f ro m tru c k to tru c k .

kF x k x x

x

k k

a) Find and write the expression for the speed, x that will lead to minimal fuel

consumption per mile of driving.

1

2

1 2

1

1 2

1 1

1 2

2

1 2

1

22

F o r th e fu le c o n s u m p tio n to b e m in im u m ,

( ) = 0

, { } 0

1, ( )+ ( ) 0

, ( )+ ( ) 0

, + 0

, + 0

, + 0

F x

kdo r k x

d x x

d do r k k x

d x x d x

d do r k x k x

d x d x

o r k x k

o r k x k

ko r k

x


1

2 2

2 1

2

1

2

,

,

,

ko r k

x

ko r x

k

ko r x

k

2 3 1

1 2 1 3

1

2

2( ) ( + ) ( 2 ) ; [+ v e , a s x is a lw a ys p o s it iv e ; c o n c a v e u p ]

T h u s w ill m a k e th e fu e l c o n s u m p tio n m in im u m

kdF x k x k k x

d x x

kx

k

b) What speed will provide minimal fuel consumption per mile of driving for a truck

having k1=4.9 and k2=0.004?

1

2

4 .93 5 m ile s p e r h o u r

0 .0 0 4

kx

k

Question 12: When a 3-ton truck is driven at a speed of x miles per hour, it travels m(x)

miles per gallon of fuel consumed, where,

2( )

5 .7 6 0 .0 0 3 6

a t w h a t s p e e d s h o u ld a tru c k b e d r iv e n i f ( ) is to b e m a x im iz e d ?

xm x

x

m x

2

2 2

2 2

2

2 2

2 2

F o r ( ) to b e m a x im u m ,

( ) = 0

, { } 05 .7 6 0 .0 0 3 6

(5 .7 6 0 .0 0 3 6 ) ( ) (5 .7 6 0 .0 0 3 6 )

, 0(5 .7 6 0 .0 0 3 6 )

(5 .7 6 0 .0 0 3 6 ) (0 .0 0 7 2 ), 0

(5 .7 6 0 .0 0 3 6 )

, 5 .7 6 0 .0 0 3 6 0 .0 0 7 2 0

, 5 .7 6 0

m x

m x

d xo r

d x x

d dx x x x

d x d xo r

x

x x xo r

x

o r x x

o r

2

2

2

.0 0 3 6 0

, 5 .7 6 0 .0 0 3 6

5 .7 6, 1 6 0 0

0 .0 0 3 6

, 4 0 m ile s p e r h o u r

x

o r x

o r x

o r x


2 2

2 2 2 2

2 2 2 2 2 2

2 4

2 2

(5 .7 6 0 .0 0 3 6 ) (0 .0 0 7 2 ) 5 .7 6 0 .0 0 3 6( ) [ ] [ ]

(5 .7 6 0 .0 0 3 6 ) (5 .7 6 0 .0 0 3 6 )

(5 .7 6 0 .0 0 3 6 ) (5 .7 6 0 .0 0 3 6 ) (5 .7 6 0 .0 0 3 6 ) (5 .7 6 0 .0 0 3 6 )

(5 .7 6 0 .0 0 3 6 )

(5 .7 6 0 .0 0 3 6 ) ( 0 .0 0 7 2

d x x x d xm x

d x x d x x

d dx x x x

d x d x

x

x

2 2

2 4

2 2 2

2 4

2 3

2 3

) 2 (5 .7 6 0 .0 0 3 6 )(5 .7 6 0 .0 0 3 6 )(0 .0 0 7 2 )

(5 .7 6 0 .0 0 3 6 )

0 .0 0 7 2 (5 .7 6 0 .0 0 3 6 ){5 .7 6 0 .0 0 3 6 2 (5 .7 6 0 .0 0 3 6 )}

(5 .7 6 0 .0 0 3 6 )

0 .0 0 7 2 {3(5 .7 6 ) 0 .0 0 3 6 } 1 2 4 4 1 6 0 .0 0 0 0 2 5 9 2

(5 .7 6 0 .0 0 3 6 ) (5 .7

x x x x

x

x x x x

x

x x x x

x

2 3

3

2 3

6 0 .0 0 3 6 )

0 .1 2 4 4 1 6 ( 4 0 ) 0 .0 0 0 0 2 5 9 2 ( 4 0 )( 4 0 ) 0 .0 0 2 1; [ , c o n c a v e d o w n ]

{5 .7 6 0 .0 0 3 6 ( 4 0 ) }

4 0 m ile s p e r h o u r m a k e s th e ( ) m a x im u m

x

m ve

x m x

Question 13: If the proportion of defective transistors in a very large stock of transistors

is P, then the proportion of good transistors is (1-P). If the probability that the first

defective is the 10th

one is C(P), where,

9( ) (1 )

w h a t va lu e o f p w ill m ax im ize ( )?

C p p p

C p

9

9 9

8 9

8 9

8

8

F o r ( ) to b e m a x im u m , ( ) 0

, { (1 ) } 0

, (1 ) (1 ) ( ) 0

, 9 (1 ) (1 ) (1 ) 0

, 9 (1 ) (1 ) 0

, (1 ) ( 9 1 ) 0

, (1 ) ( 1 0 1) 0

E ith e r , o r ,

(1 )

C p C p

do r p p

d p

d do r p p p p

d p d p

do r p p p p

d p

o r p p p

o r p p p

o r p p

p

8

0 1 0 1 0

, 1 , 0 .1

p

o r p o r p


8 8 8

8 7 8 7

7 7

7

( ) [ (1 ) ( 1 0 1)] (1 ) ( 1 0 1) ( 1 0 1) (1 )

(1 ) ( 1 0 ) 8 ( 1 0 1)(1 ) 1 0 (1 ) 8 ( 1 0 1)(1 )

(1 ) [1 0 (1 ) 8 ( 1 0 1)] (1 ) [1 0 1 0 8 0 8 ]

(1 ) (1 8 9 0 )

If 1, (1) (1

d d dC p p p p p p p

d p d p d p

p p p p p p

p p p p p p

p p

p C

7

8

8

1) (1 8 9 0 ) 0 ; [ te s t fa i l]

If 0 .5 , th e n (0 .5 ) (1 0 .5 ) { 1 0 (0 .5 ) 1)} 0 .0 1 5 ; [ v e ]

If 1 .5 , th e n (1 .5 ) (1 1 .5 ) { 1 0 (1 .5 ) 1)} 0 .0 0 0 2 1; [ v e ]

A t 1 w e h a v e a s ta t io n a ry in f le c t io n p o i n t

If 0 .1,

p C

p C

p

p

7

( 0 .1) (1 0 .1) {1 8 9 0 (0 .1)} 4 .3 0 4 ; [ v e , c o n c a v e d o w n ]

T h u s 0 .1 m a k e s ( ) m a x im u m

C

p C p

Question 14: Profit per tree grown and sold, by a tree grower depends upon the height of

a tree at the time of sale. Taking h as tree height in inches, the profit per tree, in dollars,

is approximated by, 1

2( ) (1 0 2 ) 0 .1P h h h

a) What tree height provides maximum profit per tree?

1

2

1

2

1

2

11

2

1

2

1

2

1

2

F o r ( ) to b e m a x im u m , ( ) 0

, { (1 0 2 ) 0 .1 } 0

, (1 0 2 ) ( 0 .1 ) 0

, (1 0 2 ) ( 0 .1 ) 0

1, (1 0 2 ) (1 0 2 ) 0 .1 0

2

1, (1 0 2 ) ( 2 ) 0 .1 0

2

1, 0 .1

(1 0 2 )

1, (1 0 2 )

0 .1

P h P h

do r h h

d h

d do r h h

d h d h

d do r h h

d h d h

do r h h

d h

o r h

o r

h

o r h

o

,1 0 0 1 0 2

1 0 0 1 0, 4 5 in c h e s

2

r h

o r h


1 3 3

2 2 2

3

2

1( ) { (1 0 2 ) 0 .1} (1 0 2 ) ( 2 ) (1 0 2 )

2

( 4 5 ) {1 0 2 ( 4 5 )} 0 .0 0 1; [ v e , c o n c a v e d o w n ]

T h u s , tre e h e ig h t, 4 5 in c h e s p ro v id e s m a x im u m p ro fit p e r tre e

dP h h h h

d h

P

h

b) What is the maximum profit per tree? 1

2( 4 5 ) {1 0 2 ( 4 5 )} 0 .1( 4 5 ) $ 5 .5 p e r tre eP

Question 15: Unilever Bangladesh estimates the total potential number of customers for a

new product is 1,000,000. It plans to operate a promotional campaign to sell the product

and uses the response function, 0 .0 1

( ) 0 .2 5 0 .2 5t

r t e

as a measure of the proportion of total customer potential responding to the promotion

after it has been in operation for t days. On the average, one response generates $5 in

revenue. Campaign costs consist of a fixed cost of $15,000 plus a variable cost of $1,000

per day of operation.

0 .0 1

0 .0 1

R e v e n u e 5 (1 ,0 0 0 ,0 0 0 ){ ( )} (5 , 0 0 0 , 0 0 0 ){0 .2 5 0 .2 5 }

T o ta l c o s t F ix e d c o s t + V a ria b le c o s t 1 5 0 0 0 + 1 0 0 0 t

P ro f i t R e v e n u e T o ta l c o s t (5 , 0 0 0 , 0 0 0 ){0 .2 5 0 .2 5 } (1 5 0 0 0 + 1 0 0 0 t)

t

t

r t e

e

0 .0 1

0 .0 1

1 2 5 0 0 0 0 1 2 5 0 0 0 0 1 5 0 0 0 1 0 0 0

P ro f i t P ( t ) 1 2 3 5 0 0 0 1 0 0 0 1 2 5 0 0 0 0

t

t

e t

t e

a) How long should the campaign continue if profit is to be maximized?

0 .0 1

0 .0 1

0 .0 1

0 .0 1

0 .0 1

0 .0 1

F o r p ro f i t to b e m a x im u m , ( ) 0

, {1 2 3 5 0 0 0 1 0 0 0 1 2 5 0 0 0 0 } 0

, 1 0 0 0 1 2 5 0 0 0 0 ( 0 .0 1) 0

, 1 0 0 0 1 2 5 0 0 0

,1 2 5 0 0 1 0 0 0

1 0 0 0, 0 .0 8

1 2 5 0 0

, ln ln (0 .0 8 ) ; [ ta

t

t

t

t

t

t

P t

do r t e

d t

o r e

o r e

o r e

o r e

o r e

k in g lo g o r L n in b o th s id e s ]

, 0 .0 1 ( ln ) ln (0 .0 8 ) ; [ ln lo g 1]

, 0 .0 1 ln (0 .0 8 )

ln (0 .0 8 ), 2 5 2 .5 7 3 d a y s

0 .0 1

e

eo r t e e e

o r t

o r t


0 .0 1 0 .0 1 0 .0 1

0 .0 1 ( 2 5 2 .5 7 3 ) 2 .5 2 5 7 3

( ) ( 1 0 0 0 1 2 5 0 0 ) 1 2 5 0 0 ( 0 .0 1) 1 2 5

( 2 5 2 .5 7 3) 1 2 5 1 2 5 1 0; [ v e , c o n c a v e d o w n ]

T h u s th e c a m p a ig n sh o u ld c o n tin u e fo r 2 5 2 .5 7 3 d a ys fo r th e p ro fi

t t tdP t e e e

d t

P e e

t

t to b e m a x im u m

b) What is the maximum profit? 0 .0 1 ( 2 5 2 .5 7 3 )

P (2 5 2 .5 7 3 ) 1 2 3 5 0 0 0 1 0 0 0 ( 2 5 2 .5 7 3) 1 2 5 0 0 0 0 $ 8 8 2 , 4 2 7 .1 3e

Question 16: An oil deposit contains 1,000,000 barrels of oil, which after being pumped

from the deposit, yields revenue of $12 per barrel. The proportion of the deposit that will

have been pumped out after t years of pumping is

0 .1 60 .9 0 .9

te

Operating costs are $345,600 per year.

0 .1 6

0 .1 6

0 .1 6

R e v e n u e 1 2 (1 ,0 0 0 ,0 0 0 ){0 .9 0 .9 }

O p e ra tin g c o s t 3 4 5 6 0 0 t

P ro fi t R e v e n u e c o s t 1 2 (1 ,0 0 0 ,0 0 0 ){0 .9 0 .9 } 3 4 5 6 0 0 t

P ro fi t P ( t) 1 0 8 0 0 0 0 0 1 0 8 0 0 0 0 0 3 4 5 6 0 0 t

t

t

t

e

e

e

0 .1 6

0 .1 6

0 .1 6

0 .1

) H o w lo n g s h o u ld p u m p in g b e c o n tin u e d to m a x im iz e p ro f i t ?

F o r p ro f i t to b e m a x im u m , ( ) 0

, {1 0 8 0 0 0 0 0 1 0 8 0 0 0 0 0 3 4 5 6 0 0 t} 0

, 1 0 8 0 0 0 0 0 ( 0 .1 6 ) 3 4 5 6 0 0 0

,1 7 2 8 0 0 0 3 4 5 6 0 0

,

t

t

t

P t

do r e

d t

o r e

o r e

o r e

a

6

0 .1 6

0 .1 6

3 4 5 6 0 0

1 7 2 8 0 0 0

, 0 .2

, ln ln (0 .2 ) ; [ ta k in g lo g o r L n in b o th s id e s ]

, 0 .1 6 ( ln ) ln (0 .2 ) ; [ ln lo g 1]

, 0 .1 6 ln (0 .2 )

ln (0 .2 ), 1 0 .0 6 ye a rs

0 .1 6

t

t

t

e

e

o r e

o r e

o r t e e e

o r t

o r t

0 .1 6 0 .1 6 0 .1 6

0 .1 6 (1 0 .0 6 ) 1 .6 0 9 6

( ) (1 7 2 8 0 0 0 3 4 5 6 0 0 ) 1 7 2 8 0 0 0 ( 0 .1 6 ) 2 7 6 4 8 0

(1 0 .0 6 ) 2 7 6 4 8 0 2 7 6 4 8 0 5 5 2 8 7 .0 3; [ v e , c o n c a v e d o w n ]

T h u s th e p u m p in g sh o u ld b e c o n tin u e d fo r 1 0 .0 6 ye

t t tdP t e e e

d t

P e e

t

0 .1 6 (1 0 .0 6 )

a rs fo r th e p ro fit to b e m a x im u m

) W h a t is th e m a x im u m p ro fit ?

P (1 0 .0 6 ) 1 0 8 0 0 0 0 0 1 0 8 0 0 0 0 0 3 4 5 6 0 0 (1 0 .0 6 )= $ 5 ,1 6 3 ,6 1 4 .0 8 1

b

e


Question 17: When x ounces of seed costing $2 per ounce are sown on a plot of land the

crop yield is ln(2x+1) bushels worth $25 per bushels. How many ounces should be sown if

the worth of the crop minus the cost of the seed is to be maximized?

P ro fit, P ( ) 2 5{ ln (2 1)} 2x x x

1 1 1

F o r p ro f i t to b e m a x im u m , P ( ) 0

, [ 2 5{ ln ( 2 1)} 2 ] 0

, 2 5 { ln ( 2 1)} ( 2 ) 0

1, 2 5 ( 2 ) 2 0

2 1

5 0, 2

2 1

, 2 1 2 5

, 1 2 o u n c e s o f s e e d

5 0P ( ) ( 2 ) 5 0 ( 2 1) 5 0 ( 1)( 2 1)

2 1

x

do r x x

d x

d do r x x

d x d x

o rx

o rx

o r x

o r x

d d dx x x

d x x d x d

2

2

2

( 2 1)

1 0 05 0 ( 2 1) ( 2 )

( 2 1)

1 0 0P (1 2 ) 0 .1 6 ; [ ( v e ) , c o n c a v e d o w n ]

( 2 4 1)

T h u s 1 2 o u n c e s o f s e e d w ill m a k e th e p ro f i t m a x im iz e d .

xx

xx

x

Do the followings at home

Question 18: The revenue from, and the cost of, operating an undertaking for t years are,

respectively, in millions of dollars, 0 .0 5 0 .0 8

( ) 3 a n d ( ) 1 .5t t

R t e C t e

a) How long should operations continue if profit is to be maximized?

b) Compute maximum profit.

Question 19: The total potential audience for a promotional campaign is 10,000

customers. Revenue averages $3 per response to the campaign. Campaign costs are a

fixed amount of $500, plus $300 per day the campaign continues. The proportion of the

total audience responding by time t days is

0 .2 51

te




Question 20: The total potential audience for a promotional campaign is 2,000 customers.

Revenue averages $5 per response to the campaign. Campaign costs are a fixed amount of

$100, plus $105.36 per day the campaign continues. The proportion of the total audience

responding by time t days is

1 (0 .9 )t



Marginal Propensity to Consume and the Multiplier:

The major parts of the income people have available for their use is spent on food,

clothing, shelter, medical care, transportation, recreation, and so on and this is called

consumption expenditure. The remaining amount after consumption expenditure is saved

in one form or another. Both consumption and saving depends on or function of income

(Y).

Marginal Propensity to Consume (MPC) is the rate at which consumption changes for

each additional $1 of income received. MPC is the part of each extra dollar of income that

is spent.

Marginal Propensity to Save (MPS) is the part of each extra dollar of income that is

saved.

[ ( ) ]

[ ( ) ] 1 [ ( )] 1

dC Y

d Y

d dY C Y C Y

d Y d Y

M a r g in a l P r o p e n s ity to C o n s u m e M P C

M a r g in a l P r o p e n s ity to S a v e M P S M P C

Multiplier: The rate, at which income increases for each additional $1 of investment or

savings, is called multiplier.

( )

, ( ) ( ) [ ( ) ]

( ),1 .

( ),1 [1 ]

,1 [1 ]

1,[1 ]

1 1,

[1 ]

S a v in g s I Y C Y

d d do r I Y C Y

d I d I d I

d Y d C Y d Yo r

d I d Y d I

d Y d C Yo r

d I d Y

d Yo r M P C

d I

d Yo r

M P C d I

d Yo r

d I M P C M P S

M u ltip lie r


Question 21: Suppose Y is family income in tens of thousands of dollars so that, for

example, Y=2 means $20,000 and suppose

0 .4( ) 6 0 .6

YC Y Y e

represents the family consumption function.

a) Find MPC for families with $10,000 and $20,000 of income.

0 .4 0 .4 0 .4

0 .4 (1 )

0 .4 ( 2 )

[ ( )]

(6 0 .6 ) 0 .6 ( 0 .4 ) 0 .6 0 .4

If 1; (1 0 th o u sa n d ), 0 .6 0 .4 0 .8 6 8 1

If 2 ; ( 2 0 th o u sa n d ), 0 .6 0 .4 0 .7 7 9 7

Y Y Y

dC Y

d Y

dY e e e

d Y

Y e

Y e

M a r g in a l P r o p e n s ity to C o n su m e M P C

M P C

M P C

b) Suppose Y is trillions of dollars. Find the multiplier at an income level of $0.8 trillions.

0 .4 ( 0 .8 )

If 0 .8; 0 .6 0 .4 0 .8 9 0 4

1 1 19 .1 2

[1 ] 1 0 .8 9 0 4

Y e

d Y

d I M P C M P S

M P C

M u ltip lie r


Applied Mathematics


Integral Calculus

3 3

3 1 1 1 4 2

4

2

1) (3 2 5 ) 3 2 5

3 2 5 3 2 53 1 1 1 4 2

35

4

z z m d z z d z z d z d z m d z

z z z zz m z c z m z c

zz z m z c

2

2 1 3

2 2 2

3

2 ) (7 5 )

L e t , 7 5

, (7 ) (5 ) ( )

, 7

,7

1 1 1(7 5 ) . .

7 7 7 2 1 7 3

(7 5 )

2 1

x d x

x z

d d do r x z

d x d x d x

d zo r

d x

d zo r d x

d z z zx d x z z d z c c

xc


3

3

3 1 2

3 3 3

2 2

3) (5 6 )(5 6 )

L e t , 5 6

, (5 ) (6 ) ( )

, 5

,5

1 1 1(5 6 ) . .

5 5 5 3 1 5 2

1 1

1 0 1 0 (5 6 )

d xx d x

x

x z

d d do r x z

d x d x d x

d zo r

d x

d zo r d x

d z z zx d x z z d z c c

c cz x

Problem Set 10-1:

Question No. 7:

2 2

2 1 1 1 3 2

( 2 3 4 ) 2 3 4

2 32 3 4 4

2 1 1 1 3 2

x x d x x d x x d x d x

x x x xx c x c

Question No. 8:

3 3

3 1 1 1 4 2

(3 4 1) 3 4 1

3 43 4

3 1 1 1 4 2

y y d y y d y y d y d y

y y y yy c y c

Question No. 9:

p dq p dq pq c

Question No. 10:

q dp q dp pq c


Question No. 11:

1 1 2

1 1 2

p p qp q d p q p d p q c c

Do these at home: Q.12: p q d q Q.13: 3 4

( 1) x x d x

Q.14: 2 5

( 1) y y d y Q.15: 2 4

(3 5 1) y y d y

Q.16: 3 5

( 4 6 1) x x dx Q.17: 2

x d x

Q.18: 3

7 d y

Q.19: 3

d y

y

Q.20: 4

d x

x

Question No. 21

3 1

3 3

2

2

(5 2 ) 5 2 5 23 1

15 2 5

2

yy d y d y y d y y c

yy c y c

y

Do these at home: Q.22:

4(7 3 )x dx

Q.23: 2

1( 2 1)x d x

x

Q.24: 3

2( 1)y d y

y Q.25:

1

2 p d p Q.26:

1

3 q d q Q.27:

4

33 x d x

Q.28:

3

25 y d y

Q.29: 52

3

1 2( 2 )

3

d xx

x

Q.30: 43

3

2 1( 2 )

4

x d xx

x

Question No. 31:

2 1 3 1

2 3 2 3 2 3

3 4

3 4

1 2 ( ) 1 2 ( ) 1 2[ ] 1 2[ ]2 1 3 1

1 2[ ] 4 33 4

x xx x d x x x d x x d x x d x c

x xc x x c

Do these at home: Q.32: 2

3( 2 )x x dx


Question No. 33:

3 13 3

4

4

( 2 9 ) 11 6 ( 2 9 ) 1 6 ( 2 9 ) 1 6

3 1 2

( 2 9 ) 11 6 2 ( 2 9 )

4 2

xx d x x d x c

xc x c


3 0 (3 5 )x d x Q.35: 2

(3 9 )x d x

Q.36: 4

( 2 3 )x d x

Question No. 37:

1 11

1 2 2

2

1

2

(5 3 ) 1 (5 3 ) 1(5 3 ) .( ) .

1 13 31(5 3 )

2 2

(5 3 ) 2 (5 3 )

3 3

2

d x x xx d x c c

x

x xc c

Do these at home: Q.38: 3(7 2 )

d w

w

Question No. 39:

11

1 3

3

1

3

2 2

23 3

3

8 ( 2 5 ) 18 ( 2 5 ) 8 .

1 21( 2 5 )

3

( 2 5 ) 3 ( 2 5 )4 4 6 ( 2 5 )

2 2

3

d x xx d x c

x

x xc c x c


2

1 2

(3 7 )

d x

x


Applied Mathematics


Integral Calculus

1

3

8 1

3

1

11

1 1 3

3 3

4 4

3 3

84 4

8 1 3 3

3

1

1

F in d th e a re a u n d e r ( ) 5 o v e r th e in te rv a l 1 to 8

( 5 )

H e re , ( 5 ) 5 51

13

35 5

4 4

3

3 3(8 ) 3 (1( 5 ) 5 5 (8 )

4 4

f x x x x

x d x

xx d x x d x d x x c

x xx c x c

xx d x x c c

4

3

4

3

)5 (1)

4

3 (8 ) 3 35 (8 ) 5 4 7 4 6 .2 5

4 4 4

c

c c

Problem Set 10-2:

Question No. 1:

5

5

2

2

2 2 2 (5 ) 2 ( 2 ) 6d x x c c c

Question No. 2:

3 3

3

1

1 1

3 3 3 3(3) 3(1) 6d x d x x c c c


Question No. 3:

4 44 4 1 1 2

1 1 1 1

42 2 2

1

2 2 2 21 1 2

( 4 ) ( 1) 1 5

x xx d x x d x c c

x c c c

Do this at home: Q.4:

9

3

3 x d x

Question No. 5:

6 66 6 6 1 1 2

2 2 2 2 2

62 2 2

2

( 1) 1 1 1 2

(6 ) ( 2 )(6 ) ( 2 ) 2 0

2 2 2

x xx d x x d x d x x c x c

xx c c c

Question No. 6:

33 3 3 2

32

1

1 1 1 1

2 2

( 2 1) 2 1 2 .2

(3 ) (3 ) ( 1) ( 1) 4

xx d x x d x d x x c x x c

c c

Question No. 7:

8 88

2 5 51

8 2 3 3 3

3

0

00 0

5 5

3 3

3

2 5 51

3 3

3 (8 ) 3 (0 )1 9 .2

5 5

x x xx d x c c c

c c


9 1

2

0

x d x


Question No. 9:

2 2 2 2

2 2

1 1 1 1

2 22 1 1 1 3 2

1 1

3 2 3 2

( 3 5 ) 3 5

33 . 5 5

2 1 1 1 3 2

( 2 ) 3 ( 2 ) (1) 3 (1)5 ( 2 ) 5 ( 2 ) 2 .8 3

3 2 3 2

x x d x x d x x d x d x

x x x xx c x c

c c


3

2

0

( 5 2 ) x x d x Q.12:

8 1

3

1

(1 2 ) y d y

Question No. 11:

9 9

1 11

9 9 91 1 2 2

2 2

1 1 1

1 1

91 1 1

2 2 2

1

(5 ) 5 5 51 1

12 2

5 2 5 (9 ) 2 (9 ) 5 (1) 2 (1) 4 4

y yy d y d y y d y y c y c

y y c c c

Question No. 13:

6 6

1 31

6 1 2 2

2

2

2 2

6 63 3

2 2

2 2

3 3

2 2

( 2 3 ) 1 ( 2 3 ) 1( 2 3 ) . .

1 32 21

2 2

2 ( 2 3 ) 1 ( 2 3 ).

3 2 3

{2 (6 ) 3} {2 ( 2 ) 3}8 .6 7

3 3

x xx d x c c

x xc c

c c



8 1

2

1

(5 4 ) x d x

Question No. 15:

6 6 6

2

2 2

1 1 1

6 6 62 1 1

11 1

6 0 1 6 0 6 0 (3 2 )

(3 2 ) (3 2 )

(3 2 ) 1 (3 2 ) 1 16 0 . 6 0 . 6 0

2 1 3 1 3 3(3 2 )

1 16 0

3{3(6 ) 2} 3{3(1) 2}

1 1 16 0 ( ) 6 0 (

6 0 1 5 1 5

d x d x x d xx x

x xc c c

x

c c

c c

1) 3

6 0


6

3

0 2

2 0

( 4 1)

d x

x

Q.17:

23

b

a

x d x Q.18:

34

d

c

x d x

Q.19: 1

( 1)

n

x d x Q.20:

1

( 2 1)

n

x d x

Find the area under the curve of the following functions over the given x

intervals:

Question No. 21: ( ) 2 ; 1 to 2f x x x x

22 2 1 1

22 2 2

1

1 1 1

2 2 2 . ( 2 ) (1) 31 1

xx d x x d x c x c c c


Do this at home: Q.22: ( ) 3 ; 1 to 5f x x x x

Q.23: ( ) 3 2 ; 1 to 2f x x x x Q.24: ( ) 2 3 ; 1 to 1f x x x x

Q.25: 2

6( ) ; 1 to 3f x x x

x Q.26: 3

4( ) ; 1 to 2f x x x

x

Question No. 27: 2

2

4 0( ) 4 0 ( 2 1) ; 0 to 2

( 2 1)f x x x x

x

22 2 2 1

2 2

0 0 0

2

21

0

0

( 2 1) 14 0 ( 2 1) 4 0 ( 2 1) 4 0 . .

2 1 2

2 02 0 ( 2 1)

( 2 1)

2 0 2 01 6

{2 ( 2 ) 1} {2 (0 ) 1}

xx d x x d x c

x c cx

c c

Do this at home: Q.28: 3

2

5 0( ) ; 1 to 2 5f x x x

x

Find the areas described in each of the following problems:

Question No. 29: Find the area bounded by the axis and

( ) 1 0 0 .5f x x

The point where the function touches the x –axis, f(x)=0

( ) 1 0 0 .5 0

1 0, 2 0

0 .5

f x x

o r x

The point where the function touches the y –axis, x=0

W e h a v e to f in d th e a re a u n d e r th e c u rv e w h e re ,

( ) 1 0 0 .5 ; 0 to 2 0f x x x x


2 02 0 2 0 2 0 2

0 0 0 0

2 02

0

2 2

(1 0 0 .5 ) 1 0 0 .5 1 0 0 .52

1 0 0 .2 5

1 0 ( 2 0 ) 0 .2 5 ( 2 0 ) 1 0 (0 ) 0 .2 5 (0 ) 1 0 0

xx d x d x x d x x c

x x c

c c

Question No. 30: Find the area bounded by the axis and

( ) 5f x x


( ) 5 0

, 5

f x x

o r x

The point where the function touches the y –axis, x=0

W e h a v e to f in d th e a re a u n d e r th e c u rv e w h e re ,

( ) 5 ; 0 to 5f x x x x

00 0 0 2

5 5 5 5

2 2

(5 ) 5 52

(0 ) ( 5 )5 (0 ) 5 ( 5 ) 1 2 .5

2 2

xx d x d x x d x x c

c c

Question No. 31: Find the area bounded by the x-axis and

2( ) 3 0 3f x x x


2( ) 3 0 3 0

, 3 (1 0 ) 0

E ith e r , 3 0 o r ,(1 0 ) 0

, 0 , 1 0

f x x x

o r x x

x x

o r x o r x

The function touches the x –axis, at x=0, 10


2

W e h ave to fin d th e a rea u n d er th e cu rve w h ere ,

( ) 3 0 3 ; 0 to 1 0f x x x x x

1 01 0 1 0 1 0 2 3

2 2

0 0 0 0

1 02 3 2 3 2 3

0

(3 0 3 ) 3 0 3 3 0 3 .2 3

1 5 1 5 (1 0 ) (1 0 ) 1 5 (0 ) (0 ) 5 0 0

x xx x d x x d x x d x c

x x c c c

Do these at home:

Q. 32: Find the area bounded by the x-axis and 2

( ) 4 2 1f x x x

Q. 33: Find the area bounded by the x-axis and 3 2

( ) 3 3f x x x x

Q. 34: Find the area bounded by the x-axis and 4 2

( ) 5 4f x x x


Applied Mathematics


Integral Calculus

1

1

1( )

( )

L e t ,

, ( ) ( ) ( )

,

,

1 1 1 1 1( ) ( ln )

( )

1ln ( )

m x b d x d xm x b

m x b z

d d do r m x b z

d x d x d x

d zo r m

d x

d zo r d x

m

d zm x b d x d x d z z c

m x b z m m z m

m x b cm

33 3

3

1

11 1

4 1 1 4 4 . ln ( 2 3 ) 2 ln ( 2 3 )

2 3 2 3 2

2 ln (9 ) 2 ln (1) 2 ln (9 ) 2 ln (1) 4 .3 9

d xd x x c x c

x x

c c

33 3

3

1

11 1

6 1 1 6 6 . ln (3 2 ) 2 ln (3 2 )

3 2 3 2 3

2 ln (7 ) 2 ln (1) 2 ln (7 ) 2 ln (1) 3 .8 9

d xd x x c x c

x x

c c


Problem Set 10-5:

Question No. 1:

1 1lnx d x d x x c

x

Question No. 2:

1ln

d xd x x c

x x

Question No. 3:

2 12 2 ln

d xd x x c

x x


1 13 3 3 lnx d x d x x c

x

Question No. 5:

2 1 1

2 1 1

2 1 1

x xx d x c c x c c

x

Question No. 6:

2 1 1

2 1

2

1

2 1 1

d x x xx d x c c x c c

x x

Question No. 7:

1 1ln (5 4 )

5 4 5 4 5

d xd x x c

x x

Question No. 8

1ln (3 )

3d x x c

x

Question No. 9:

1 1ln (3 0 .2 ) 5 ln (3 0 .2 )

3 0 .2 0 .2d x x c x c

x


Question No. 10:

1 1 12 (0 .5 1) 2 2 ln (0 .5 1)

(0 .5 1) 0 .5

4 ln (0 .5 1)

x d x d x x cx

x c

Question No. 11:

2 1

2

2

1

1 ( 2 1)( 2 1) .

( 2 1) 2 2 1

1 1( 2 1)

2 2 ( 2 1)

d x xx d x c

x

x c cx

Question No. 12:

2 1

2

2

1

1 0 1 (5 3)1 0 (5 3) 1 0 .

(5 3) 5 2 1

22 (5 3)

(5 3)

d x xx d x c

x

x c cx

Question No. 13:

1 0 1 0

1 0

1

1 1

1 ln ln (1 0 ) ln (1) 2 .3 0

d xd x x c c c

x x

Question No. 14:

1

1

1 1

1 ln ln ( ) ln (1) 1

e e

e

x d x d x x c e c cx

Question No. 15:

22

2

0

00

1ln (0 .5 4 ) 2 ln (0 .5 4 )

0 .5 4 0 .5

2 ln (5 ) 2 ln ( 4 ) 0 .4 4 6

d xx c x c

x

c c


1 0

5

1( )

0 .6 1d x

x


( )

( ) ( )

1 01 0 1 0

0 .0 6 0 .0 6 0 .0 6

00 0

1 00 .0 6 0 .0 6 (1 0 ) 0 .0 6 ( 0 )

0

1

13 0 3 0 3 0 .

0 .0 6

5 0 0 5 0 0 5 0 0

4 1 1 .0 5

x x

m x b

m x b m x b

x x x

x

e d x e c

ee d x e c c

m m

e d x e d x e c

e c e c e c

2 22

2 1 2 1 2 1

1 11

3 1

1 1.

2 2

1 18 .6 8 3

2 2

ln

1.

ln ln

x x x

x

x

m x b m x b

m x b

e d x e c e c

e c e c

aa d x c

a

a aa d x c c

m a m a

22 2 1

2 1

1 1

3 1

(0 .9 ) 1(0 .9 ) .

ln (0 .9 ) 2

(0 .9 ) (0 .9 )0 .8 1 1

2 ln (0 .9 ) 2 ln (0 .9 )

x

xd x c

c c


4 44 0 .5 0 .5

0 .5

2 2 2

2 1

(6 ) 1 2 (6 )(6 ) .

ln (6 ) 0 .5 ln (6 )

2 (6 ) 2 (6 )3 3 .4 8 7

ln (6 ) ln (6 )

x x

xd x c c

c c

Problem Set 10-6:

Question No. 1:

x xe d x e c

Question No. 2:

22

ln 2

x

xd x c

Question No. 3:

33

ln 3

x

xd x c

Question No. 4:

x xe d x e c

Question No. 5:

0 .5 0 .5 0 .512

0 .5

x x xe d x e c e c

Question No. 6:

0 .2 0 .2 1 0 .2

0 .2 1 5 5 55 . 5

0 .2 ln 5 ln 5 ln 5

x x x

xd x c c c

Question No. 7:

1 0 .4 1 0 .4

1 0 .4 1 ( 0 .5 ) ( 0 .5 )( 0 .5 ) . 2 .5

0 .4 ( ln 0 .5 ) ( ln 0 .5 )

x x

xd x c c


Question No. 8:

2 0 .5 2 0 .5 2 0 .51. 2

0 .5

x x xe d x e c e c

Question No. 9:

3 0 .1 3 0 .1 3 0 .112 2 2 0

0 .1

x x xe d x e c e c

Question No. 10:

5 0 .2 5 0 .2 5 0 .214 4 2 0

0 .2

x x xe d x e c e c

Question No. 11:

( 0 .8 ) 1( 0 .8 )

( 0 .8 ) ( ln 0 .8 ) ( 0 .8 ) ( ln 0 .8 )

x

x

x x

d xd x c c

Question No. 12:

0 .5 0 .5

0 .5 0 .5

1 1 2

0 .5

x x

x xd x e d x e c c

e e

Question No. 13:

555

1 0 .2 1 0 .2 1 0 .2

0

00

1 1 1

12 2 . 1 0

0 .2

1 0 1 0 1 7 .1 8 2

x x xe d x e c e c

e c e c

Question No. 14:

222

2 0 .5 2 0 .5 2 0 .5

1

11

1 1 .5

14 4 . 8

0 .5

8 8 1 4 .1 0 7

x x xe d x e c e c

e c e c

Question No. 15:

22

1 1

2 1

(0 .5 )1 0 (0 .5 ) 1 0

ln (0 .5 )

(0 .5 ) (0 .5 )1 0 1 0 3 .6 0 6

ln (0 .5 ) ln (0 .5 )

x

xd x c

c c

Question No. 16:


22

0 0

2 0

(0 .9 )5 (0 .9 ) 5

ln (0 .9 )

(0 .9 ) (0 .9 )5 5 9 .0 1 7

ln (0 .9 ) ln (0 .9 )

x

xd x c

c c

Problem Set 10-7:

Question No. 1:

2

2 2.2 2 { ( ) 2 }

ln 2 ln 2

2 1 2 1 2 22 . { (ln 2 ) 1}

ln 2 ln 2 ln 2 ln 2 ln 2 (ln 2 )

x x

x x x

x x x x

x

dx d x x d x x d x d x x d x

d x

x xd x c x c

Question No. 2:

6

2 1

L e t, 2 1

, ( 2 ) (1) ( )

, 2

,2

xd x

x

x z

d d do r x z

d x d x d x

d zo r

d x

d zo r d x

a g a in ,

2 1

, 2 1

, 6 3 3

x z

o r x z

o r x z

6 1 1 1 16 (3 3 ) (3 3 )

2 1 2 1 2 2

1 3 1 3 1(3 ) [ 3 ] [3 3 ln ]

2 2 2

3 3 3 3ln ( 2 1) ln ( 2 1)

2 2 2 2

3 3 33 ln ( 2 1) 3 ln ( 2 1)

2 2 2

x d zd x x d x z z d z

x x z z

d z d z d z z z cz z

z z c x x c

x x c x x c


Question No. 3:

2

2

( 0 .5 1)

L e t , 0 .5 1

, ( 0 .5 ) (1) ( )

, 0 .5

1,

2

, 2

xd x

x

x z

d d do r x z

d x d x d x

d zo r

d x

d zo r

d x

o r d x d z

a g a in ,

0 .5 1

, 0 .5 1

, 2 4 4

x z

o r x z

o r x z

2 2 2 2

2 1

2

2 1 1 12 ( 4 4 ) 2 (8 8 )

(0 .5 1) (0 .5 1)

8 18 8 ln 8 8 ln 8

2 1

8 18 ln (0 .5 1) 8[ ln (0 .5 1)]

(0 .5 1) (0 .5 1)

xd x x d x z d z z d z

x x z z

zd z z d z z c z c

z z

x c x cx x

Question No. 4:

. { ( ) }

( 1)

x x x x x

x x x

dx e d x x e d x x e d x d x x e e d x

d x

x e e c e x c

Question No. 5:

2 0 .5 2 0 .5 2 0 .5

2 0 .5 2 0 .5 2 0 .5 2 0 .5

2 0 .5 2 0 .5 2 0 .5

. { ( ) }

1 1 12 2

0 .5 0 .5 0 .5

2 4 2 ( 2 )

x x x

x x x x

x x x

dx e d x x e d x x e d x d x

d x

xe e d x xe e c

xe e c e x c


Question No. 6:

2

9

(3 4 )

L e t , 3 4

, (3 ) ( 4 ) ( )

, 3

1,

3

xd x

x

x z

d d do r x z

d x d x d x

d zo r

d x

o r d x d z

a g a in ,

3 4

, 3 4

, 9 3 ( 4 )

, 9 3 1 2

x z

o r x z

o r x z

o r x z

2 2 2 2

2 1

2 2

9 1 1 19 (3 1 2 ) ( 4 )

(3 4 ) (3 4 ) 3

1 4 1 1 1( ) 4 ln 4 ln 4

2 1

4ln (3 4 )

(3 4 )

x d zd x x d x z z d z

x x z z

zd z d z d z z c z c

z z z z z

x cx

Question No. 7:

2

(0 .5 1)

2 (0 .5 1 0 .5 ) 2 (0 .5 1) 2 (0 .5 ){ } { }

(0 .5 1) (0 .5 1) (0 .5 1)

2 1 2 1{ }

(0 .5 1) (0 .5 1)

12 ln ln (0 .5 1) 2 ln 2 ln (0 .5 1)

0 .5

2{ ln ln (0 .5 1)} 2 ln(0 .5 1)

d xx x

x x x xd x d x

x x x x x x

d x d x d xx x x x

x x c x x c

xx x c c

x


Question No. 8:

0 .4 1 0 .4 1 0 .4 1

0 .4 1 0 .4 1 0 .4 1 0 .4 1

0 .4 1 0 .4 1

2 . 2 { ( 2 ) }

1 2 52 5

0 .4 0 .4 0 .4

5 1 2 .5

x x x

x x x x

x x

dx e d x x e d x x e d x d x

d x

x e e d x x e e c

x e e c

Question No. 9:

2

5 3

L e t , 5 3

, (5 ) (3 ) ( )

, 5

,5

xd x

x

x z

d d do r x z

d x d x d x

d zo r

d x

d zo r d x

a g a in ,

5 3

, 5 3

3,

5

2 6, 2

5

x z

o r x z

zo r x

zo r x

2 1 1 2 6 1 12 . . ( 2 6 )

5 3 5 3 5 5 2 5

1 6 1 6 1( 2 ) { 2 } {2 6 ln }

2 5 2 5 2 5

1 2 3 6{2 (5 3 ) 6 ln (5 3 )} ln (5 3 )

2 5 5 2 5 2 5

2 6 3ln (5 3 ) 2[ ln (5 3 )]

5 2 5 5 2 5

x z d zd x x d x z d z

x x z z

d z d z d z z z cz z

x x c x x c

xx x c x c

Question No. 10:

6

(3 2 )

3(3 2 ) 9 ) 3(3 2 ) 9{ } { }

(3 2 ) (3 2 ) (3 2 )

d xx x

x x x xd x d x

x x x x x x


3 9 3 9 1{ } 3 ln 9 . ln (3 2 )

(3 2 ) (3 2 ) 3

3 ln 3 ln (3 2 ) 3[ln ln (3 2 )] 3 ln(3 2 )

d x d x d x x x cx x x x

xx x c x x c c

x

Question No. 11:

20 .1 2

2

2

.

L e t , 0 .1 2

, (0 .1 ) ( 2 ) ( )

, 0 .2

, 50 .2

xx e d x

x z

d d do r x z

d x d x d x

d zo r x

d x

d zo r x d x d z

2 20 .1 2 0 .1 2

. 5 5 5x z z x

x e d x e d z e c e c

Question No. 12:

0 .2

0 .2

0 .2

0 .2

0 .2

1

L e t, 1

, (1) ( ) ( )

, 0 .2

,0 .2 0 .2 ( 1)

x

x

x

x

x

d x

e

e z

d d do r e z

d x d x d x

d zo r e

d x

d z d zo r d x

e z


Applied Mathematics


Integral Calculus

2

F in d th e a re a b o u n d e d b y th e fu n c tio n s

( ) 1 5 2 a n d

( ) 9

A re a , [ ( ) ( ) ]

T h e v a lu e s o f a a n d b a re th e x -c o o rd in a te s o f th e p o in ts o f

in te rs e c t io n o f ( ) a n d ( )

A t th e p o in t o f in te rs

b

a

f x x x

g x x

A f x g x d x

f x g x

2

2

2

2

2

2

2

3

2 2 1 1 1

2

3

e c tio n , ( ) ( )

,1 5 2 9

,1 5 2 9 0

, 6 0

, 6 0

, 3 2 6 0

, ( 2 )( 3 ) 0

, 2 , 3

A re a , [ ( ) ( ) ] [1 5 2 (9 )]

[ 6 ] 62 1 1 1

b

a

f x g x

o r x x x

o r x x x

o r x x

o r x x

o r x x x

o r x x

o r x

A f x g x d x x x x d x

x xx x d x x c

2 23 2

3 3

3 2 3 2

3 2 3 2

63 2

2 2 ( 3 ) ( 3 )1 2 6 ( 3 )

3 2 3 2

2 2 ( 3 ) ( 3 )1 2 6 ( 3 ) 2 0 .8 4

3 2 3 2

x xx c

c c

c c


3

F in d th e a re a b e tw e e n th e c u rv e s

( ) 3 3 a n d

( )

A re a , [ ( ) ( ) ]



A t th e p o in t o f in te rs e c t io n , (

b

a

f x x x

g x x

A f x g x d x

f x g x

f

3

3

2

2

2

) ( )

, 3 3

, 3 4 0

, (3 4 ) 0

E ith e r , o r ,

0 (3 4 ) 0

4 o r ,

3

2 o r ,

3

2 2, , 0 ,

3 3

, 1

x g x

o r x x x

o r x x

o r x x

x x

x

x

o r x

o r x

1 .1 5 5

3

0

1 .1 5 5 1 .1 5 53 1 1 1 4

2

0 0

4 4

2 2

.1 5 5 , 0 , 1 .1 5 5

A re a , [ ( ) ( ) ] 2 [3 4 ]

32 3 4 2 2

3 1 1 1 4

3 (1 .1 5 5 ) 3 (0 )2{ 2 (1 .1 5 5 ) 2 (0 ) }

4 4

2 ( 1 .3 3 3 4 ) 2 .6 7

A re a c a n n o t b e

b

a

A f x g x d x x x d x

x x xc x c

c c

n e g a t iv e , h e n c e a re a 2 .6 7


Problem Set 10-3

Question No:3

2


( ) 1 a n d

( ) 1 0

A re a , [ ( ) ( ) ]



A t th e p o in t o f in te rs e c t io

b

a

f x x

g x

A f x g x d x

f x g x

2

2

33 2 1

2

3 3

33 3 3

3

n , ( ) ( )

, 1 1 0

, 9 0

, 3

A re a , [ ( ) ( ) ] [ 9 ] 92 1

3 ( 3 )9 9 (3 ) 9 ( 3 )

3 3 3

3 6

A re a c a n n o t b e n e g a tiv e , h e n c e A re a 3 6

b

a

f x g x

o r x

o r x

o r x

xA f x g x d x x d x x c

xx c c c

Do this at home: Question No:4

2


( ) 3 4 a n d

( ) 9

f x x

g x


Question No:5

2

F in d th e a re a b o u n d e d b y th e fu n c t io n s

( ) 8 2 0 a n d

( ) 1 4

A re a , [ ( ) ( ) ]



A t th e p o in t o f in te r

b

a

f x x x

g x x

A f x g x d x

f x g x

2

2

2

2

6

2

1

6 62 1 1 1 3 2

1 1

3

s e c t io n , ( ) ( )

, 8 2 0 1 4

, 8 2 0 1 4 0

, 7 6 0

, 6 6 0

, ( 6 )( 1) 0

, 6 , 1

A re a , [ ( ) ( ) ] [ 7 6 ]

77 6 6

2 1 1 1 3 2

(6 )

3

b

a

f x g x

o r x x x

o r x x x

o r x x

o r x x x

o r x x

o r x


x x x xx c x c

2 3 27 (6 ) (1) 7 (1)

6 (6 ) 6 (1)2 3 2

2 0 .8 4

A re a c a n n o t b e n e g a t iv e , h e n c e A re a 2 0 .8 4

c c


3


( ) a n d

( )

f x x

g x x


Question No:6

2

F in d th e a re a b o u n d e d b y th e fu n c t io n s

( ) 2 0 2 a n d

( ) 1 2 2

A re a , [ ( ) ( ) ]



A t th e p o in t o f in te r

b

a

f x x

g x x x

A f x g x d x

f x g x

2

2

2

2

2

5

2

2

52 1 1 1

2

s e c t io n , ( ) ( )

, 2 0 2 1 2 2

, 2 0 2 1 2 2 0

, 2 1 4 2 0 0

, 2 ( 7 1 0 ) 0

, 5 2 1 0 0

, ( 5 )( 2 ) 0

, 5 , 2

A re a , [ ( ) ( ) ] [ 2 1 4 2 0 ]

22 1 4 2 0

2 1 1 1

b

a

f x g x

o r x x x

o r x x x

o r x x

o r x x

o r x x x

o r x x

o r x


x xx c

53

2

2

3 3

2 2

7 2 03

2 (5 ) 2 ( 2 )7 (5 ) 2 0 (5 ) 7 ( 2 ) 2 0 ( 2 )

3 3

9

A re a c a n n o t b e n e g a t iv e , h e n c e A re a 9

xx x c

c c


4 2

F in d th e a rea b o u n d ed b y th e fu n c tio n s

( ) 5 + 4 an d ( ) 4f x x x g x


Applied Mathematics

Applications of Integral Calculus

Question1: The total amount of coal a country will consume in a period of years depends

upon the rate of consumption, and this rate increases as time, t years increases. Suppose it

is estimated that the consumption rate, t years from now will be,

( ) 2 0 1 .2 m illio n to n s p e r yearr t t

a) Compute the total amount of coal the country will consume in the next 10 years.

1 01 0 1 1

0 0

1 02

0

2 2

m illio n to n s[ 2 0 1 .2 ] ( ye a r) 2 0 1 .2 m illio n to n s

ye a r 1 1

2 0 0 .6 m illio n to n s

2 0 (1 0 ) 0 .6 (1 0 ) 2 0 (0 ) 0 .6 (0 ) m illio n to n s

2 6 0 m illio n to n s

tt d t t c

t t c

c c

b) How much coal will be consumed in the following 10 years, that is during the

second decade from now?

2 02 0

2

1 0

1 0

2 2

[ 2 0 1 .2 ] 2 0 0 .6 m illio n to n s

2 0 ( 2 0 ) 0 .6 ( 2 0 ) 2 0 (1 0 ) 0 .6 (1 0 ) m illio n to n s

3 8 0 m illio n to n s

t d t t t c

c c

c) If the total supply of coal available to the country now and in the future is 2,500

million tons, how long will it be until the total supply is exhausted?

Let us assume, we can use the total supply (2,500 million tons) of coal up to X

years from now.

0

2

0

2 2

2

2

[ 2 0 1 .2 ] 2 5 0 0

, 2 0 0 .6 2 5 0 0

, 2 0 ( ) 0 .6 ( ) 2 0 (0 ) 0 .6 (0 ) 2 5 0 0

, 2 0 0 .6 2 5 0 0

, 3 1 0 0 1 2 5 0 0 ; [m u lt ip ly in g e a c h s id e b y 5 ]

x

x

t d t

o r t t c

o r x x c c

o r x x

o r x x


2, 3 2 5 0 1 5 0 1 2 5 0 0 0

, (3 2 5 0 ) 5 0 (3 2 5 0 ) 0

, (3 2 5 0 )( 5 0 ) 0

2 5 0, , 5 0

3

Y e a rs c a n n o t b e n e g a tiv e , h e n c e , 5 0

T h e s u p p ly w ill b e e x h a u s te d in 5 0 ye a rs

o r x x x

o r x x x

o r x x

o r x y ea r s y ea r s

x y ea r s

Question2: Maintenance cost on a new machine is at the rate of 100t dollars per year at

time t years.

( ) 1 0 0 d o lla rs p e r yearc t t

a) Compute total maintenance cost for the first 4 years.

44 1 1

0 0

42 2 2

0

d o lla rs[1 0 0 ] ( y e a r) 1 0 0 d o lla rs

ye a r 1 1

5 0 5 0 ( 4 ) 5 0 (0 ) $ 8 0 0

tt d t c

t c c c

b) How many years will it take for total maintenance cost to amount to $5,000?

Let us assume, it will take up to X years from now for total maintenance cost to

amount to $5,000.

0

2

0

2 2

2

2

[1 0 0 ] 5 0 0 0

, 5 0 5 0 0 0

, 5 0 ( ) 5 0 (0 ) 5 0 0 0

, 5 0 ( ) 5 0 0 0

, 1 0 0

, 1 0 ye a rs

x

x

t d t

o r t c

o r x c c

o r x

o r x

o r x

It will take up to 10 years from now for total maintenance cost to amount to $5,000.


Question3: The fixed cost incurred when g gallons of paint are produced is $2,500 and

marginal cost at g gallons of output is

2( ) 0 .0 0 0 4 5 0 .1 8 2 0c g g g

Find the total cost of producing 100 gallons

1 0 0

2

0

1 0 02 1 1 1

0

1 0 03 2

0

3 2

T o ta l C o s t F ix e d c o s t [0 .0 0 0 4 5 0 .1 8 2 0 ]

$ 2 5 0 0 0 .0 0 0 4 5 0 .1 8 2 02 1 1 1

$ 2 5 0 0 0 .0 0 0 1 5 0 .0 9 2 0

$ 2 5 0 0 0 .0 0 0 1 5 (1 0 0 ) 0 .0 9 (1 0 0 ) 2 0 (1 0 0 )

g g d g

g gg c

g g g c

c

3 2 0 .0 0 0 1 5 (0 ) 0 .0 9 (0 ) 2 0 (0 )

$ 2 5 0 0 $ 1 2 5 0 $ 3 7 5 0

c

Question4: Marginal cost at output level p pounds of nails is,

1

2( ) 1 ( 1)c p p

and fixed cost is $400. Find the total cost of making 399 pounds of the product.

3 9 9 1

2

0

3 9 9

13 9 91

12

2

0

0

1 1

2 2

T o ta l C o s t F ix e d c o s t [1 ( 1) ]

( 1)$ 4 0 0 $ 4 0 0 2 ( 1)

11

2

$ 4 0 0 3 9 9 2 (3 9 9 1) 0 2 (0 1)

$ 4 0 0 $ 4 3 7 $ 8 3 7

p d p

pp p p c

c c


Question5: Suppose that an operation provided a company income at the rate of ( )I t

dollars per day at time t days, where,

1

2( ) 1 1 0 4I t t

1

2( ) 2 0 7E t t

The operation was started at an initial fixed expense of $2,000. Variable expense at the

rate of ( )E t per day is incurred at time t days. From a profit maximization viewpoint, the

operation should be continued as long as income per day ( )I t exceeds expense per day

( )I t . Calculate net maximum profit.

1 1

2 2

1 1

2 2

1

2

1

2

2

9 0 0 9 0 0

0 0

9 0 0

0

9 0 0 1

2

0

P ro fi t is m a x im u m ti ll,

( ) ( )

,1 1 0 4 2 0 7

, 7 4 9 0

, 3 9 0

, 3 0

, 3 0 9 0 0

N e t P ro f i t [ ( ) ] [ ( ) ] F ix e d c o s t

[ ( ) ( ) ] $ 2 0 0 0

[1 1 0 4 2 0

I t E t

o r t t

o r t t

o r t

o r t

o r t d a y s

I t d t E t d t

I t E t d t

t

1

2

9 0 0

11

9 0 0 1 2

2

0

0

9 0 01 .5

0

1 .5 1 .5

7 ] $ 2 0 0 0

[9 0 3 ] $ 2 0 0 0 9 0 3 $ 2 0 0 01

12

9 0 2 $ 2 0 0 0

9 0 (9 0 0 ) 2 (9 0 0 ) 9 0 (0 ) 2 (0 ) $ 2 0 0 0

$ 2 5 , 0 0 0

t d t

tt d t t c

t t c

c c


Question6: Maintenance cost of newly purchased equipment is expected to cost ( 2 0 .1 )t

thousand dollars per year at time t years.

a) Compute total maintenance cost during the first 6 years.

66 1 1

62

0

0 0

2 2

[ ( 2 0 .1 )] 2 0 .1 2 0 .0 51 1

2 (6 ) 0 .0 5 (6 ) 2 (0 ) 0 .0 5 (0 )

$ 1 3 .8 th o u s a n d d o lla rs

= $ 1 3 .8 (1 0 0 0 )= $ 1 3 ,8 0 0

tt d t t c t t c

c c

b) Compute total maintenance cost during the second 6 years.

1 21 2 1 1

1 22

6

6 6

2 2

[ ( 2 0 .1 )] 2 0 .1 2 0 .0 51 1

2 (1 2 ) 0 .0 5 (1 2 ) 2 (6 ) 0 .0 5 (6 )

$ 1 7 .4 th o u s a n d d o lla rs

= $ 1 7 .4 (1 0 0 0 )= $ 1 7 ,4 0 0

tt d t t c t t c

c c

c) At what time t, will the total spent on maintenance cost reach $60 thousand?

0

1 1

0

2

0

2 2

2

2

2

[ ( 2 0 .1 )] 6 0

, 2 0 .1 6 01 1

, 2 0 .0 5 6 0

, 2 ( ) 0 .0 5 ( ) 2 (0 ) 0 .0 5 (0 ) 6 0

, 2 0 .0 5 6 0

, 0 .0 5 2 6 0 0

, 0 .0 5 3 6 0 0

, 0 .0 5 ( 6 0 ) 1( 6 0 ) 0

, ( 6 0 )(

t

t

t

t d t

to r t c

o r t t c

o r t t c c

o r t t

o r t t

o r t t t

o r t t t

o r t

0 .0 5 1) 0

E ith e r , o r ,

6 0 0 0 .0 5 1 0

, 6 0 o r , 2 0

ye a rs c a n n o t b e n e g a t iv e , h e n c e a f te r 2 0 ye a rs th e

to ta l m a in te n a n c e c o s t w

t

t t

o r t t

t

i ll re a c h $ 6 0 th o u s a n d .


Question7: A currently new product is expected to be sold at the rate of ( 2 0 0 .4 )t

thousand dollars per year at time t years, t<50.

a) Compute total sales during the first 10 years.

1 01 0 1 1

1 02

0

0 0

2 2

[ ( 2 0 0 .4 )] 2 0 0 .4 2 0 0 .21 1

2 0 (1 0 ) 0 .2 (1 0 ) 2 0 (0 ) 0 .2 (0 )

$ 1 8 0 th o u s a n d d o lla rs

= $ 1 8 0 (1 0 0 0 )= $ 1 8 0 0 0 0

tt d t t c t t c

c c

b) Compute total sales during the second 10 years.

2 02 0 1 1

2 02

1 0

1 0 1 0

2 2

[ ( 2 0 0 .4 )] 2 0 0 .4 2 0 0 .21 1

2 0 ( 2 0 ) 0 .2 ( 2 0 ) 2 0 (1 0 ) 0 .2 (1 0 )

$ 1 4 0 th o u s a n d d o lla rs

= $ 1 4 0 (1 0 0 0 )= $ 1 4 0 0 0 0

tt d t t c t t c

c c

c) At what time t, will total sales reach $375 thousand?

0

1 1

0

2

0

2 2

2

2

2

[ ( 2 0 0 .4 )] 3 7 5

, 2 0 0 .4 3 7 51 1

, 2 0 0 .2 3 7 5

, 2 0 0 .2 2 0 (0 ) 0 .2 ( 0 ) 3 7 5

, 2 0 0 .2 3 7 5

, 0 .2 2 0 3 7 5 0

, 0 .2 1 5 5 3 7 5 0

, 0 .2 ( 7 5 ) 5 ( 7 5 ) 0

, (

t

t

t

t d t

to r t c

o r t t c

o r t t c c

o r t t

o r t t

o r t t t

o r t t t

o r t

7 5 )(0 .2 5 ) 0

E ith e r , o r ,

7 5 0 0 .2 5 0

, 7 5 o r , 2 5

W e w ill ta k e th e s m a lle s t f ig u re b e c a u s e w e w a n t to a c h ie v e s a le s ta rg e t

a t

t

t t

o r t t

e a r lie s t p o s s ib le t im e , h e n c e a f te r 2 5 y e a rs s a le s w ill re a c h $ 3 7 5 th o u s a n d .


Question8: An industry consumes fuel at rate of1 / 2

( 2 0 .6 )t million barrels per year at

time t years. How much fuel will the industry consume in 25 years.

2 5

11

2 52

2 51 / 2 1 .5

0

0

0

1 .5 1 .5

[ ( 2 0 .6 )] 2 0 .6 2 0 .41

12

2 ( 2 5 ) 0 .4 ( 2 5 ) 2 (0 ) 0 .4 (0 )

1 0 0 m illio n b a rre ls

tt d t t c t t c

c c

Question9: An oil rig pumps oil from a well at the rate of 1 / 2

(3 6 0 7 2 )t barrels per year

at time t years. How much oil will be pumped in the next 9 years.

9

11

92

91 / 2 1 .5

0

0

0

1 .5 1 .5

[ (3 6 0 7 2 )] 3 6 0 7 2 3 6 0 4 81

12

3 6 0 (9 ) 4 8 (9 ) 3 6 0 (0 ) 4 8 (0 )

1 9 4 4 b a rre ls

tt d t t c t t c

c c

Question10: At time t years, an industry consumes fuel at the rate of 1 / 2

( 2 9 )t million

barrels per year. If the total supply of fuel available to the industry now and in the future

is 63 million barrels, how many years will the supply last.

1 / 2

0

11

2

0

1 .5

0

1 .5 1 .5

[ ( 2 9 ) ] 6 3

1 ( 2 9 ), . 6 3

121

2

( 2 9 ), 6 3

3

( 2 9 ) (9 ), 6 3

3 3

t

t

t

t d t

to r c

to r c

to r c c


1 .5

1 .5

1 11 .5

1 .5 1 .5

( 2 9 ), 9 6 3

3

, ( 2 9 ) 2 1 6

, ( 2 9 ) 2 1 6

, 2 9 3 6

, 1 3 .5 ye a rs

to r

o r t

o r t

o r t

o r t

Question11: At time t years, sales of a currently new product are expected to be

1 / 2

1 / 2

1 01 0 (0 .5 1 6 )

(0 .5 1 6 )t

t

million dollars per year. How many years will it take for total sales to amount to $40

million?

1 / 2

0

11

2

0

0 .5

0

0 .5 0 .5

0 .5 0 .5

0 .5

0 .

[1 0 (0 .5 1 6 ) ] 4 0

(0 .5 1 6 ), 1 0 .2 . 4 0

11

2

, 4 0 (0 .5 1 6 ) 4 0

, 4 0 (0 .5 1 6 ) 4 0 (1 6 ) 4 0

, 4 0 (0 .5 1 6 ) 4 0 (1 6 ) 4 0

, (0 .5 1 6 ) 5

, (0 .5 1 6 )

t

t

t

t d t

to r c

o r t c

o r t c c

o r t

o r t

o r t

5 2 25

, 0 .5 1 6 2 5

, 1 8 ye a rs

x

o r t

o r t

Question12: The population of a trading area is currently 100 thousand. At time t years

from now population will be growing at the rate of

1 / 2

1 / 2

2 02 0 (0 .5 9 )

(0 .5 9 )t

t

thousand per year. What will total population be 14 years from now?


1 4

11

1 42

1 / 2

0

0

1 40 .5

0

0 .5 0 .5

(0 .5 9 )[ 2 0 (0 .5 9 ) ] 2 0 .2 .

11

2

8 0 (0 .5 9 )

8 0{0 .5 (1 4 ) 9} 8 0{0 .5 (0 ) 9} 8 0

P o p u la t io n w ill in c re a s e 8 0 m illio n f ro m n o w .

T h e re fo re a f te r 1 4 ye a r

tt d t c

t c

c c

s p o p u la t io n w ill b e 1 0 0 8 0 1 8 0 m illio n

Question13: A fixed cost of $2 thousand has been incurred in setting up an advertising

campaign. Variable cost is expected to be at the rate of 1 / 2

(3 0 .0 6 )t thousand dollars per

month at time t months. Estimate total cost if the campaign runs 25 months.

2 5

1 / 2

0

2 5

11

2 52

2 51 / 2 1 .5

0

0

0

1 .5 1 .5

T o ta l c o s t F ix e d c o s t [ (3 0 .0 6 )]

( )[ (3 0 .0 6 )] 3 0 .0 6 3 0 .0 4 ( )

11

2

3( 2 5 ) 0 .0 4 ( 2 5 ) 3 (0 ) 0 .0 4 (0 ) 8 0

T o ta l c o s t 2 8 0 $ 8 2 th o u s a n d

t d t

tt d t t c t t c

c c

Question14: When t tons of steel are produced, marginal cost in dollars per ton is 2

(0 .0 0 6 1 .2 5 0 )t t . If fixed cost is $1,600, find the total cost of producing 100 tons.

1 0 0

2

0

1 0 01 0 0 2 1 1 1

2

0 0

1 0 03 2

0

3 2 3 2

T o ta l c o s t F ix e d c o s t [ (0 .0 0 6 1 .2 5 0 )]

( )[ (0 .0 0 6 1 .2 5 0 )] 0 .0 0 6 1 .2 5 0

2 1 1 1

0 .0 0 2 0 .6 5 0

0 .0 0 2 (1 0 0 ) 0 .6 (1 0 0 ) 5 0 (1 0 0 ) 0 .0 0 2 (0 ) 0 .6 (0 ) 5

t t d t

t tt t d t t c

t t t c

c

0 (0 )

$ 1 0 0 0

T o ta l c o s t $ 1 6 0 0 $ 1 0 0 0 $ 2 6 0 0

c


Do this at home: Question15: When b barrels of whiskey are produced, marginal cost in

dollars per barrel is2

(0 .0 0 4 5 1 .8 2 0 0 )b b . If fixed cost is $4,000, find the total cost of

producing 200 barrels.

Question16: A fixed cost of $50 thousand was incurred in setting up an operation. At time

t months thereafter, the operation yields income at the rate of ( 2 0 0 .3 )t and incurs

expense at the rate of (1 0 0 .1 )t where both rates are in thousands of dollars per month.

a) What is the optimal time to terminate the operation?

Profit will be maximum till income per day exceeds expenses per day.

( 2 0 0 .3 ) (1 0 0 .1 )

, 0 .2 1 0

, 5 0 ye a rs

t t

o r t

o r t

Optimal time to terminate the operation is 50 years

b) What will total profit be at the optimal time of termination?

5 0 5 0

0 0

5 0 5 0

0 0

5 05 0 1 1

0 0

P ro f i t T o ta l in c o m e o p e ra t in g c o s t F ix e d c o s t

[ ( 2 0 0 .3 )] [ (1 0 0 .1 )] 5 0

[ ( 2 0 0 .3 ) (1 0 0 .1 )] 5 0 [ ( 2 0 0 .3 1 0 0 .1 ] 5 0

[ (1 0 0 .2 ] 5 0 1 0 0 .2 5 0 1 0 0 .11 1

t d t t d t

t t d t t t d t

tt d t t c t t

5 0

2

0

2 2

5 0

1 0 (5 0 ) 0 .1(5 0 ) 1 0 (0 ) 0 .1(0 ) 5 0

$ 2 0 0 th o u s a n d

c

c c

Do this at home: Question17: A fixed cost of $4,100 was incurred in setting up an

operation. At time t months thereafter, the operation yields income at the rate of

1 / 2( 2 0 0 0 1 0 0 )t and incurs expense at the rate of

1 / 2( 2 0 0 0 2 0 0 )t where both rates

are in dollars per month.

a) What is the optimal time to terminate the operation?

b) What will total profit be at the optimal time of termination?


Question18: If new reserves of fuel are discovered at the rate of

1 0 0

0 .2 1t

million barrels per year at time t years, find the total amount of fuel that will be

discovered in the next 25 years.

Answer:

2 52 5 2 5

00 0

2 5

0

1 0 0 1 1[ ] [1 0 0 ] 1 0 0 . ln (0 .2 1)

0 .2 1 0 .2 1 0 .2

5 0 0 ln (0 .2 1)

5 0 0 ln (5 1) 5 0 0 ln (1) 8 9 5 .8 8 m illio n b a rre ls

d t d t t ct t

t c

c c

Question19: A company projects its cost of providing medical care to workers to be at the

rate of 0 .0 3

1 5t

e thousand of dollars per year at time t years.

a) Compute total medical care cost for the next 10 years.

1 01 01 0

0 .0 3 0 .0 3 0 .0 3

0

00

0 .0 3 (1 0 ) 0 .0 3 ( 0 )

1[1 5 ] 1 5 5 0 0

0 .0 3

5 0 0 5 0 0 $ 1 7 4 .9 3 th o u sa n d

t t te d t e c e c

e c e c

b) How long will it be until total cost amounts to $250 thousand?

0 .0 3

0

0 .0 3

0

0 .0 3

0

0 .0 3 0 .0 3 ( 0 )

0 .0 3

0 .0 3

0 .0 3

[1 5 ] 2 5 0

1, 1 5 2 5 0

0 .0 3

, 5 0 0 2 5 0

, 5 0 0 5 0 0 2 5 0

, 5 0 0 5 0 0 2 5 0

, 1 .5

, ln ln (1 .5 )

, 0 .0 3 ln ln (1 .5 )

, 0 .0 3 ln (1 .5 )

t

t

t

t

tt

t

t

t

t

e d t

o r e c

o r e c

o r e c e c

o r e

o r e

o r e

o r t e

o r t

ln (1 .5 ), 1 3 .5 1 y e a rs

0 .0 3o r t


Do this at home: Question20: The total supply of a fuel available now and in the future is

1,000 million barrels. At time t years from now, fuel will be consumed at the rate of

0 .0 51 0

te million barrels per year.

a) How much fuel will be consumed in the next 20 years.

b) How long will the supply of fuel last?

Do this at home: Question21: At time t years, the cost of maintaining a facility is at the

rate of 0 .0 8

1 2t

e thousands of dollars per year.

a) Find total maintenance cost for the next 10 years.

b) How long will it take for total maintenance cost to reach $300 thousand?

Do this at home: Question22: At time t years, interest on a bank account is at the rate of

0 .0 66 0 0

te dollars per year.

a) What will be the total interest accumulation in 12 years?

b) How long will it take for total interest accumulation to reach $5,000?

Do this at home: Question23: Sales of a product are projected to be at the rate of

0 .21 5

te

million pounds per year at time t years.

a) Find total sales in the next 5 years?

b) How long will it take for total sales to reach 60 million pounds?

Question24: Sales of wheat at time t years are projected to be at the rate of 0 .2

5 1 5t

e

million pounds per year. Find total sales in the next 5 years?

555

0 .2 0 .2 0 .2

0

00

0 .2 ( 5 ) 0 .2 ( 0 )

1[5 1 5 ] 5 1 5 5 7 5

0 .2

5 (5 ) 7 5 5 (0 ) 7 5 7 2 .4 1 m illio n p o u n d s

t t te d t t e c t e c

e c e c

Do this at home: Question25: Sales of milk at time t years are projected to be at the rate

of 0 .4

1 0 2 0t

e

million gallons per year. Find total sales in the next 5 years?


Consumers’ Surplus: Consumers’ Surplus is the amount of dollars left in the hands of

consumers due to the market settling at an equilibrium point because of competition; it is

the difference between what consumers expected to pay and what they actually paid.

0

[ ( )]

mq

d m mP q d q p q

Question26: At market equilibrium, consumers demand 625,000 gallons of kerosene,

which has the demand function,

1

2( ) 2 5 0 .6d

P q q

Where q is in thousands of gallons and ( )d

P q is price of kerosene in dollars per gallon.

Compute consumers’ surplus.

Answer: First we determine the equilibrium point (qm, pm).

At market equilibrium, consumers demand qm = 625 thousand gallons

1

22 5 0 .6 ( 6 2 5 ) $ 1 0 p e r g a llo nm

P

Price per unit

Pd(q)

Consumers’

Surplus

E (qm, pm)

Demand

Curve

q

qm

0


6 2 5 1

2

0

6 2 5

16 2 51

32

2

0

0

1 .5 1 .5

T h e n c o n s u m e rs ' S u rp lu s = [ 2 5 0 .6 ] 1 0 (6 2 5 )

2 5 0 .6 6 2 5 0 2 5 0 .4 6 2 5 01

12

2 5 (6 2 5 ) 0 .4 (6 2 5 ) 2 5 (0 ) 0 .4 (0 ) 6 2 5 0

$ 3 1 2 5 th o u s a n d

= $ 3,1 2 5 , 0 0 0

q d q

qq c q q c

c c

Question27: At market equilibrium, consumers demand 100,000 tons of SAE 90

lubricating oil, which has the demand function,

( ) 1 1 0 0 .5d

P q q

Where q is in thousands of tons and ( )d

P q is price of SAE 90 lubricating oil in dollars per

ton. Compute consumers’ surplus.


At market equilibrium, consumers demand qm = 100 thousand tons

1 1 0 0 .5 (1 0 0 ) $ 6 0 p e r to nm

P

1 0 0

0

1 0 01 1

1 0 02

0

0

2 2

T h e n c o n s u m e rs ' S u rp lu s = [1 1 0 0 .5 ] 6 0 (1 0 0 )

1 1 0 0 .5 6 0 0 0 1 1 0 0 .2 5 6 0 0 01 1

1 1 0 (1 0 0 ) 0 .2 5 (1 0 0 ) 1 1 0 (0 ) 0 .2 5 (0 ) 6 0 0 0

$ 2 5 0 0 th o u s a n d

= $ 2 , 5 0 0 , 0 0 0

q d q

qq c q q c

c c


Do this at home: Question28: The demand function for a product is

( ) 7 5 0 .6d

P q q where q is in millions of barrels and ( )d

P q is price of the

product in dollars per barrel. Market equilibrium occurs at a demand of 100 million

barrels. Compute consumers’ surplus.


2

8 0( )

(0 .1 0 .2 )d

P qq

where q is in millions of tons and ( )d

P q is price of the product in dollars per ton. Market

equilibrium occurs at a demand of 18 million tons. Compute consumers’ surplus.


5 0( )

0 .5 1d

P qq

where q is in millions of pounds and ( )d

P q is price of the product in dollars per pound.

Market equilibrium occurs at a demand of 18 million pounds. Compute consumers’

surplus.


Producers’ Surplus: Producers’ Surplus is the amount of additional dollars received by

producers due to the market settling at an equilibrium point because of competition; it is

the difference between what producers actually receive and what they expected to receive.

0

[ ( )]

mq

m m sp q P q d q

Question31: At market equilibrium, consumers demand 625,000 gallons of kerosene,

where the supply function,

1

2( ) 2 .5 0 .3s

P q q

Where q is in thousands of gallons and ( )s

P q is price of kerosene in dollars per gallon.

Compute producers’ surplus.


At market equilibrium, consumers demand=Producer’ supply= qm = 625 thousand

gallons

1

22 .5 0 .3 ( 6 2 5 ) $ 1 0 p e r g a llo nm

P

Price per unit

Ps (q)

Producers’

Surplus

E (qm, pm)

Supply

Curve

q

qm

0


6 2 5 1

2

0

6 2 5

16 2 51

32

2

0

0

1 .5 1 .5

T h e n p ro d u c e rs ' S u rp lu s =1 0 (6 2 5 ) [ 2 .5 0 .3 ]

1 0 (6 2 5 ) 2 .5 0 .3 6 2 5 0 2 .5 0 .21

12

6 2 5 0 2 .5 (6 2 5 ) 0 .2 (6 2 5 ) 2 .5 (0 ) 0 .2 (0 )

$ 1 5 6 2 .5 th o u s a n d

= $ 1, 5 6

q d q

qq c q q c

c c

2 , 5 0 0

Do this at home: Question32: The supply function for a product is

( ) 5 0 .2s

P q q

where q is in millions of tons and ( )s

P q is price of the product in dollars per ton. Market

equilibrium occurs at a demand of 50 million tons. Compute producers’ surplus.

Do this at home: Question33: The supply function for a product is

3

2( ) ( 4 0 .2 )s

P q q

where q is in thousands of truckloads and ( )s

P q is price of the product in dollars per

truckload. Market equilibrium occurs at a demand of 60 thousand truckloads. Compute

producers’ surplus.

Do this at home: Question34: The supply function for a product is0 .0 2

( ) 5q

sP q e

where q is in thousands of pounds and ( )s

P q is price of the product in dollars per pound.

Market equilibrium occurs at a demand of 40 thousand pounds. Compute producers’

surplus.

Do this at home: Question35: The supply function for a product is0 .0 5

( ) 1 0 2q

sP q e

where q is in thousands of gallons and ( )s

P q is price of the product in dollars per gallon.

Market equilibrium occurs at a demand of 20 thousand gallons. Compute producers’

surplus.


Question36: The supply and demand functions for SAE 90 lubricating oil are

( ) 1 0 0 .5s

P q q and ( ) 1 1 0 0 .5d

P q q ,where q is in thousands of tons and ( )P q

is price of SAE 90 lubricating oil in dollars per ton.

a) Find the equilibrium point

A t e q u ilib r iu m p o in t , D e m a n d = S u p p ly

i .e . ( ) ( )

,1 1 0 0 .5 1 0 0 .5

, 0 .5 0 .5 1 1 0 1 0

, 1 0 0 th o u s a n d s o f to n s

E q u ilib r iu m P ric e , 1 1 0 0 .5 (1 0 0 ) $ 6 0 p e r to n

E q u ilib r iu m p o in t ( , )= (1 0 0 , 6 0 )

d s

m

m m

P q P q

o r q q

o r q q

o r q

P

q p

b) Compute consumers’ surplus.

1 0 0

0

1 0 01 1

1 0 02

0

0

2 2

T h e n c o n s u m e rs ' S u rp lu s = [1 1 0 0 .5 ] 6 0 (1 0 0 )

1 1 0 0 .5 6 0 0 0 1 1 0 0 .2 5 6 0 0 01 1

1 1 0 (1 0 0 ) 0 .2 5 (1 0 0 ) 1 1 0 (0 ) 0 .2 5 (0 ) 6 0 0 0

$ 2 5 0 0 th o u s a n d

= $ 2 , 5 0 0 , 0 0 0

q d q

qq c q q c

c c

c) Compute producers’ surplus.

1 0 0

0

1 0 01 1

1 0 02

0

0

2 2

T h e n P ro d u c e rs ' S u rp lu s = 6 0 (1 0 0 ) [1 0 0 .5 ]

6 0 0 0 1 0 0 .5 6 0 0 0 1 0 0 .2 51 1

6 0 0 0 1 0 (1 0 0 ) 0 .2 5 (1 0 0 ) 1 0 (0 ) 0 .2 5 (0 )

$ 2 5 0 0 th o u s a n d

= $ 2 , 5 0 0 , 0 0 0

q d q

qq c q q c

c c


Do this at home: Question37: The supply and demand functions for a product are

( ) 1 0 0 .1s

P q q and ( ) 1 1 0 0 .2d

P q q ,where q is in thousands of tons and ( )P q

is price of the product in dollars per ton.




Do this at home: Question38: The supply and demand functions for a product are

( ) 1 0 .0 2s

P q q and ( ) 6 0 .0 8d

P q q ,where q is in millions of pounds and ( )P q

is price of the product in dollars per pound.





Applied Mathematics

Matrices and its applications

Problem set: 2-4

Question9:

3 1 2 1 5 2 3 1 1 5 2 2

1 4 1 3 2 4 1 3 4 2 1 4

4 4 0

2 6 5

Question11:

1 3 2 2 5 3 4 1 2 8 6 1 5 94 3

5 1 3 1 2 1 2 0 4 1 2 3 6 3

4 6 1 2 1 5 8 9 2 2 7 1 7

2 0 3 4 6 1 2 3 1 7 2 9

Question12:

1 2 3 0 5( 2 2 )( 2 3 ) ( 2 3 )

3 5 1 2 0

1 3 2 1 5 1 0 2 2 4 1 5 2 0 5

3 3 5 1 1 4 3 0 5 2 1 0 3 5 5 0 1 5

5 4 5

1 4 1 0 1 5

x x x

x x x x x x

x x x x x x

Question13:

1 2

5 6 7 0 3 (1 3 )(3 2 ) (1 2 )

3 1

5 1 6 0 7 3 2 6 5 2 6 3 7 1 3 5

2 6 3 5

x x x

x x x x x x


Question21: Interest at the rates 0.06, 0.07 and 0.08 is earned on respective investments of

$3,000, $2,000 and $4,000.

a) Express the total amount of interest earned as the product of a row vector and a

column vector.

3 0 0 0

0 .0 6 0 .0 7 0 .0 8 2 0 0 0

4 0 0 0

b) Compute the total interest by matrix multiplication.

3 0 0 0

0 .0 6 0 .0 7 0 .0 8 2 0 0 0 0 .0 6 3 0 0 0 0 .0 7 2 0 0 0 0 .0 8 4 0 0 0

4 0 0 0

$ 6 4 0

x x x

Question22: Two canned meat spreads, Regular and Superior, are made by grinding

beef, pork, and lamb together. The numbers of pounds of each meat in a 15 pound batch

of each brand are as follows:

Brand Pounds of

Beef Pork Lamb

Superior 8 2 5

Regular 4 8 3

a) Suppose we wish to make 10 batches of Superior and 20 of Regular. Multiply the

meat matrix in the table and the batch vector (10 20) and interpret the result.

8 2 51 0 2 0

4 8 3

1 0 8 2 0 4 1 6 0 1 0 2 2 0 8 1 8 0 1 0 5 2 0 3 1 1 0

1 6 0 1 8 0 1 1 0

x x x x x x

We will need 160 pounds of Beef, 180 pounds pork and 110 pounds of lamb to

process all those bathces.


b) Suppose that the per pound prices of beef, pork, and lamb are $2.50, $2.00 and

$3.00, respectively. Multiply the price vector and the meat matrix and interpret

the results.

8 2 52 .5 2 3 (C o s t x T yp e o f M e a t ) (B ra n d x T yp e o f M e a t)

4 8 3

(1 3 )( 2 3 ) N o t p o s s ib le

8 4

2 .5 2 3 2 8 (C o s t x T yp e o f M e a t ) (T yp e o f M e a t x B ra n d )

5 3

(1 3 )(3 2 ) (1

x x

x x

x 2 ) (C o s t x B ra n d )

2 .5 8 2 2 3 5 3 9 2 .5 4 2 8 3 3 3 5

3 9 3 5

x x x x x x

The total cost of a batch of Superior is $39 and a batch of Regular is $35

Question23: Three jewellery salespeople each regularly buy a certain quantity of watches,

rings, necklaces, and ear-rings from their respective wholesale supplier and sell certain

quantities of each item to a retailer. Their respective buying and selling transactions

during a particular month are given by the following vectors:

1 2 3

1 2 3

1 0 2 0 3 0 4 0 1 5 2 5 3 5 4 5 2 0 1 0 4 5 3 5

8 2 0 2 5 3 5 1 5 2 0 3 0 4 0 1 5 1 0 4 0 3 5

B B B

S S S

where the first entry in each vector represents the number of watches, the second entry

the number of rings, the third entry the number of necklaces, the forth entry the number

of sets of earrings. The buying prices were $50 per watch, $40 per ring, $25 per necklace,

and $10 per set of earrings; whereas the selling prices were, respectively, $75, $55, $30,

and $15.

a) Write the buying prices as a column vector B and the selling prices as a column

vector S.

5 0 7 5

4 0 5 5

2 5 3 0

1 0 1 5

B S

b) Find the amount of revenue each salesperson expended in buying.


5 01 0 2 0 3 0 4 0 1 0 5 0 2 0 4 0 3 0 2 5 4 0 1 0 2 4 5 0

4 01 5 2 5 3 5 4 5 1 5 5 0 2 5 4 0 3 5 2 5 4 5 1 0 3 0 7 5

2 52 0 1 0 4 5 3 5 2 0 5 0 1 0 4 0 4 5 2 5 3 5 1 0 2 8 7 5

1 0

2 4 5 0

3 0 7 5

2 8 7 5

x x x x

x x x x

x x x x

Salespeople expended $2450, $3075 and $2875 respectively in buying.

c) Find the amount of revenue each salesperson collected in selling.

7 58 2 0 2 5 3 5 8 7 5 2 0 5 5 2 5 3 0 3 5 1 5 2 9 7 5

5 51 5 2 0 3 0 4 0 1 5 7 5 2 0 5 5 3 0 3 0 4 0 1 5 3 7 2 5

3 01 5 1 0 4 0 3 5 1 5 7 5 1 0 5 5 4 0 3 0 3 5 1 5 3 4 0 0

1 5

2 9 7 5

3 7 2 5

3 4 0 0

x x x x

x x x x

x x x x

Salespeople collected $2975, $3725 and $3400 respectively in selling.

d) Find the amount of profit each salesperson realized (assuming, for simplicity, that

the unsold jewellery can not be returned or otherwise sold).

2 9 7 5 2 4 5 0 2 9 7 5 2 4 5 0 5 2 5

P ro f i t 3 7 2 5 3 0 7 5 3 7 2 5 3 0 7 5 6 5 0

3 4 0 0 2 8 7 5 3 4 0 0 2 8 7 5 5 2 5

Salespeople realized $525, $650 and $525 profit respectively.

Determinant

Determinant is a value associated with a square matrix.

1 2D e te rm in a n t o f is

3 5

1 5 3 2 1

A

A x x

1 3 3D e te rm in a n t o f is

4 1

1 3 1 3 4 1

B

B x x

http://en.wikipedia.org/wiki/Square_matrix#Square_matrices


2 3 1

D e te rm in a n t o f 1 4 2 is

5 6 4

2{ ( 4 4 ) (6 2 )} 3{ (1 4 ) (5 2 )} 1{ (1 6 ) ( 4 5 )} 1 2

A

A x x x x x x

Inverse Matrix

Gauss-Jordan Method:

1 1 2

2 2 1

1

3 2F in d th e in v e rs e o f

1 1

3 2 1 0 1 0 1 2A u g m e n te d m a tr ix , [ ( 2 )

1 1 0 1 1 1 0 1

1 0 1 2[ ]

0 1 1 3

1 2

1 3

A

A I R R R

R R R

A

1 1 2

2 2 1

1


2 1

7 3 1 0 1 0 1 3A u g m e n te d m a tr ix , [ ( 3 )]

2 1 0 1 2 1 0 1

1 0 1 3[ ( 2 )]

0 1 2 7

1 3

2 7

A

A I R R R

R R R

A


1 2 2 1

1 1 1 1 2

1


2 3

0 1 1 0 2 3 0 1A u g m e n te d m a tr ix , [ , ]

2 3 0 1 0 1 1 0

1 1 .5 0 0 .5 1 0 1 .5 0 .5[ / 2 ] [ ( 1 .5 )]

0 1 1 0 0 1 1 0

1 .5 0 .5

1 0

A

A I R R R R

R R R R R

A

Cofactor and Ad-joint Matrix Method:

0 1F in d th e in ve rse o f

2 3A

Step 1: Find the determinant

0 1D ete rm in an t o f , 0 3 2 1 2

2 3A D x x

Step 2: Find the cofactors and cofactor matrix:

1 1 1 2

2 1 2 2

A A

A A

1 1 1 2

2 1 2 2

3 2

1 0

A A

A A

1

3 2T h e c o fa c to r m a tr ix , A

1 0

S te p 3 : F in d th e a d jo in t m a tr ix ( t ra n s p o s e o f c o fa c to r m a tr ix )

3 1T h e a d jo in t m a tr ix , A A

2 0

S te p 4 : F in d th e in v e rs e m a tr ix

1 1T h e in v e rs e m a tr ix , A A

2

c

Tc

a d j

Tc

D

3 13 1

2 22 0

1 0


Find the inverse matrix of

2 3 1

1 4 2

5 6 4


2{ (4 4 ) (6 2 )} 3{ (1 4 ) (5 2 )} 1{ (1 6 ) (4 5)} 1 2D x x x x x x


1 1 1 2 1 3

2 1 2 2 2 3

3 1 3 2 3 3

A A A

A A A

A A A

1 1 1 2 1 3

2 1 2 2 2 3

3 1 3 2 3 3

4 2 1 2 1 44 ( 6 ) 6 1 4

6 4 5 4 5 6

3 1 2 1 2 36 3 ( 3 ) 3

6 4 5 4 5 6

3 1 2 1 2 32 3 5

4 2 1 2 1 4

4 6

T h e c o fa c to r m a tr ix , Ac

A A A

A A A

A A A

1 4

6 3 3

2 3 5

1

S te p 3 : F in d th e a d jo in t m a tr ix ( t ra n s p o s e o f c o fa c to r m a tr ix )

4 6 2

T h e a d jo in t m a tr ix , A A 6 3 3

1 4 3 5


4 6 2

1 1T h e in v e rs e m a tr ix , A A 6 3 3

1 21 4 3 5

Tc

a d j

Tc

D

1 1 1

3 2 6

1 1 1

2 4 4

7 1 5

6 4 1 2


Find the inverse matrix of

3 2 1

1 1 0

2 0 1


3(1 0 ) 2 ( 1 0 ) 1(2 ) 3D


1 1 1 2 1 3

2 1 2 2 2 3

3 1 3 2 3 3

A A A

A A A

A A A

1 1 1 2 1 3

2 1 2 2 2 3

3 1 3 2 3 3

1 0 1 0 1 11 1 2

0 1 2 1 2 0

2 1 3 1 3 22 1 4

0 1 2 1 2 0

2 1 3 1 3 21 1 5

1 0 1 0 1 1

1 1 2

T h e c o fa c to r m a tr ix , Ac

A A A

A A A

A A A

1

2 1 4

1 1 5

S te p 3 : F in d th e a d jo in t m a tr ix ( tra n s p o s e o f c o fa c to r m a tr ix )

1 2 1


2 4 5


1 2 1

1 1T h e in v e rs e m a tr ix , A A 1

3

Tc

a d j

Tc

D

1 2 1

3 3 3

1 1 11 1

3 3 32 4 5

2 4 5

3 3 3


Find the inverse matrix of 1)

7 3

2 1

2) 2 2

3 5

3) 9 4

2 1

4) 1 1

1 2

5) 1 3

2 0

6) 2 5

3 4

7) 0 1

2 3

8)

0 3

2 5

9)

2 8 1 1

1 5 7

1 2 3

10)

2 2 3

0 1 1

4 0 3

11)

1 1 1

1 1 1

2 2 2

12)

2 1 4

3 0 2

1 2 3

Answers: 1)

1 3

2 7

2)

5 1

4 2

3 1

4 2

3) 1 4

2 9

4)

2 1

1 1

5)

10

2

1 1

3 6

6)

4 5

7 7

3 2

7 7

7)

3 1

2 2

1 0

8)

5 1

6 2

10

3

9)

1 2 1

4 5 3

3 4 2

10)

3 13

2 2

2 3 1

2 4 1

11) inverse does not exist 12)

4 5 2

9 9 9

7 2 8

9 9 9

2 1 1

3 3 3

Find the solution:

1 2

1 2

1

2

1

2

2 3 1 7

2 1 0

W e c a n w rite th e a b o v e e q u a tio n s in th e m a tr ix fo rm a s fo llo w s :

2 3 1 7

1 2 1 0

2 3 1 7,

1 2 1 0

x x

x x

x

x

xo r

x


1 1 2

2 2 1

1 1 2

1

2

1 2

2 3 1 7 1 1 7[ ]

1 2 1 0 1 2 1 0

1 1 7[ ]

0 1 3

1 0 4[ ]

0 1 3

4

3

. . 4 , 3

R R R

R R R

R R R

x

x

i e x x

Find the solution:

1 2

1 2

1

2

1

2

1 1 2

2 2

7 3 5

2 7

W e c a n w ri te th e a b o v e e q u a t io n s in th e m a tr ix fo rm a s fo llo w s :

7 3 5

2 1 7

7 3 5,

2 1 7

7 3 5 1 0 1 6[ ( 3 )]

2 1 7 2 1 7

1 0 1 6[

0 1 3 9

x x

x x

x

x

xo r

x

R R R

R R

1

1

2

1 2

( 2 )]

1 6

3 9

. . 1 6 , 3 9

R

x

x

i e x x


Find the solution:

1 2 3

2 3

1 2 3

1

2

3

1

2

3

2 2 3 3

2

4


2 2 3 3

0 1 1 2

1 1 1 4

2 2 3 3

, 0 1 1 2

1 1 1 4

x x x

x x

x x x

x

x

x

x

o r x

x

1 1 3

3 1 3

2 2 3

1 1 2

1 1 3

2 2 3 3 1 1 2 1

0 1 1 2 0 1 1 2 [ ]

1 1 1 4 1 1 1 4

1 1 2 1

0 1 1 2 [ ]

0 0 1 5

1 1 2 1

0 1 0 7 [ ]

0 0 1 5

1 0 2 8

0 1 0 7 [ ]

0 0 1 5

1 0 0 2

0 1 0 7 [

0 0 1 5

R R R

R R R

R R R

R R R

R R R

( 2 )]


1

2

3

1 2 3

2

7

5

. . 2 , 7 , 5

x

x

x

i e x x x

Question: Suppose a company makes liquid products in three grades (Lowgrade,

Midgrade, and Highgrade) which contain different amounts of additives, A1, A2 and A3

per gallon as shown in the following table:

Liquid Gallons

Made

Pounds of Additive per Gallon

A1 A2 A3

Lowgrade 1

x 1 1 1

Midgrade 2

x 1 2 1

Highgrade 3

x 2 0 1

20 30 20

1 2 3

1 2 3

1 2 3

2 2 0

2 0 3 0

2 0

x x x

x x x

x x x

1 1

2 2

3 3


1 1 2 2 0 1 1 2 2 0

1 2 0 3 0 , h e re , 1 2 0 , , 3 0

1 1 1 2 0 1 1 1 2 0

, .

x x

x A X x B

x x

o r A X B

1

, .

1 1 2

F irs t w e n e e d to f in d th e in v e rs e m a tr ix o f 1 2 0

1 1 1

o r X A B

A



1(2 0 ) 1(1 0 ) 2 (1 2 ) 1D


1 1 1 2 1 3

2 1 2 2 2 3

3 1 3 2 3 3

2 0 1 0 1 22 1 1

1 1 1 1 1 1

1 2 1 2 1 11 1 0

1 1 1 1 1 1

1 2 1 2 1 14 2 1

2 0 1 0 1 2

2 1 1

T h e c o fa c to r m a tr ix , A 1 1 0

4 2 1

S te p 3 : F in d th e a d jo i

c

A A A

A A A

A A A

1

n t m a tr ix ( t ra n s p o s e o f c o fa c to r m a tr ix )

2 1 4


1 0 1


2 1 4 2 1 4

1T h e in v e rs e m a tr ix , A A 1 1 1 2 1 1 2

1 0 1 1 0 1

Tc

a d j

Tc

D

1

1

1

2

3

1 2 3

2 1 4 2 0 2 2 0 1 3 0 4 2 0 1 0

A . 1 1 2 3 0 1 2 0 1 3 0 2 2 0 1 0

1 0 1 2 0 1 2 0 0 3 0 1 2 0 0

1 0

A . 1 0

0

1 0 , 1 0 0

x x x

B x x x

x x x

x

X x B

x

x x x

10 gallons of Lowgrade, 10 gallons of Midgrade, and 0 gallons of Highgrade product will

be produced.


Using Cramer’s Rule:

1 2 3

1 2 3

1 2 3

2 2 0

2 0 3 0

2 0

x x x

x x x

x x x

1 1 2

1 2 0 1( 2 0 ) 1(1 0 ) 2 (1 2 ) 1

1 1 1

D

1

2 0 1 2

3 0 2 0 2 0 ( 2 0 ) 1(3 0 0 ) 2 (3 0 4 0 ) 1 0

2 0 1 1

D x

2

1 2 0 2

1 3 0 0 1(3 0 0 ) 2 0 (1 0 ) 2 ( 2 0 3 0 ) 1 0

1 2 0 1

D x

3

1 1 2 0

1 2 3 0 1( 4 0 3 0 ) 1( 2 0 3 0 ) 2 0 (1 2 ) 0

1 1 2 0

D x

1

1

2

2

3

3

1 01 0

1

1 01 0

1

00

1

D xx

D

D xx

D

D xx

D


Find the solution:

1 2

1 2

8 5 2

3 2 1

x x

x x

8 51 6 1 5 1

3 2D

1

2 54 5 1

1 2D x

2

8 28 6 2

3 1D x

1 2

1 2

1 21 2

1 1

D x D xx x

D D

Find the solution for the following (using Gauss-Jordan Method, Inverse Matrix Method,

and Cramer’s rule method)

1)

1 2

1 2

4 3 2

9 7 3

x x

x x

2)

1 2

1 2

6 8 3

2 3 1

x x

x x

3)

1 2 3

1 2 3

1 2 3

3 0 5 3

2 2 5 7

0 2

x x x

x x x

x x x

4)

1 2

2 3

1 2 3

7 3 1

3 5 2

3

x x

x x

x x x

5)

1 2 3

1 2 3

1 3

2 3

2 0

3 2 3

x x x

x x x

x x

Answers: Q1) 1 2( , ) (5 , 6 )x x Q2) 1 2

( , ) (0 .5 , 0 )x x

Q3) 1 2 3( , , ) (6 , 5 , 3 )x x x Q4) 1 2 3

( , , ) (3 7 , 8 6 , 5 2 )x x x

Q5) 1 2 3( , , ) ( 3, 0 , 6 )x x x


Applied Mathematics

Linear Programming and Simplex Method

Question1: A steel manufacturing company produces two different types of steel products

in three manufacturing divisions. The following table illustrates per unit time required in

each manufacturing division, the available hours in each division and the contribution

per unit of each product:

Manufacturing

Division

Product X

Hours per

unit

Product Y

Hours per

unit

Available

Hours

Constraints

A 8 10 11,000 8 1 0 1 1, 0 0 0x y

B 4 10 9,000 4 1 0 9 , 0 0 0x y

C 12 6 12,000 1 2 6 1 2 , 0 0 0x y

Contribution/unit $4 $8

Find the optimal production plan (maximize contribution) for the company.

Objective function, maximize contribution, 4 8z x y

Drawing Graph:

Let us assume an iso-contribution line of contribution=$3200

x y

0 1100

1375 0

x y

0 900

2250 0

x y

0 2000

1000 0

x y

0 400

800 0

x y

0 2000

1000 0

Division B

Division C

Division A

E

F

G

H X

Y

Contribution

Line

Feasible

Region


The optimum point is found by drawing an example of an iso-contribution line on the

diagram (any value of Z will do) and then placing a ruler against it. Then, by moving the

ruler away from the origin (in the case of a maximization problem) or towards the origin

(in the case of a minimization problem) by keeping it parallel to the iso-contribution line,

the last corner it touches of the feasible region represents the optimum solution.

The optimum corner is F, the intersection of Division A and Division B.

8 1 0 1 1, 0 0 0

4 1 0 9 , 0 0 0

, 4 2 0 0 0

, 5 0 0

, 7 0 0

x y

x y

o r x

o r x

a n d y

The optimum production plan is to produce 500 units of product X and 700 units of

product Y.

The maximum contribution we can get =4x+8y=$7,600

Shadow Prices and Slack:

Slack is the amount by which a resource is under-utilized. It will occur when the optimum

point does not fall on a given resource line. Slack is important because unused resources

can be allocated for other use.

The shadow price of a resource can be found by calculating the increase in value (extra

contribution or extra profit) which would be created by having available one additional

unit of a limiting resource at its original cost.

Shadow price therefore represents the maximum premium that the firm should be willing

to pay for one extra unit of each constraint.

The optimum corner F is found in the intersection of Division A and Division B.

Therefore hours available at Division A and B are fully utilized and are referred to as

critical constraints. We can calculate shadow prices for the critical constraints.

However there is unutilized hours available at Division C which is referred to as non-

critical constraint and can be allocated for other use. Non-critical constraints will have

zero shadow prices as slack exists already.


Shadow price calculation:

Division A and B is Critical constraint.

Shadow price for Division A:

Adding one extra hour for Division A, while leaving the other critical constraint

unchanged.

8 1 0 1 1, 0 0 1

4 1 0 9 , 0 0 0

, 4 2 0 0 1

, 5 0 0 .2 5

, 6 9 9 .9

x y

x y

o r x

o r x

a n d y

The increased contribution =4x500.25+8x699.9—$7,600=$0.2

The increase of $0.20 is the shadow price or the premium that the company would be

willing to pay for each extra hour in Division A.

Similarly we can calculate shadow price for Division B. There is no shadow price for

Division C as slack is available there.

Microsoft Solver Outputs:


Objective Cell (Max)

Cell Name

Original Value Final Value

$C$2 Max. Contribution z 0 7600

Variable Cells

Cell Name

Original Value Final Value Integer

$A$2 x 0 500 Contin

$B$2 y 0 700 Contin

Constraints

Cell Name Cell Value Formula Status Slack

$B$11 Non-neg 500 $B$11>=0

Not Binding 500

$B$13 Non-neg 700 $B$13>=0

Not Binding 700

$B$5 Division A 11000 $B$5<=11000 Binding 0

$B$7 Division B 9000 $B$7<=9000 Binding 0

$B$9 Division C 10200 $B$9<=12000

Not Binding 1800

Binding: Critical Constraints (resource or hours fully utilized)

Non-binding: Non-critical constraints (Division C: slack- 1800 hours available or no

shadow price)

Constraints

Final Shadow Constraint Allowable Allowable

Cell Name Value Price R.H. Side Increase Decrease

$B$11 Non-neg 500 0 0 500 1E+30

$B$13 Non-neg 700 0 0 700 1E+30

$B$5 Division A 11000 0.2 11000 750 2000

$B$7 Division B 9000 0.6 9000 2000 1000

$B$9 Division C 10200 0 12000 1E+30 1800

Shadow price for Division A=$0.2 and Division B=$0.6

Variable

Lower Objective

Upper Objective

Cell Name Value

Limit Result

Limit Result

$A$2 x 500

0 5600

500 7600

$B$2 y 700

0 2000

700 7600


Simplex Method

Objective function, maximize contribution, 4 8z x y

Constraints:

8 1 0 1 1, 0 0 0

4 1 0 9 , 0 0 0

1 2 6 1 2 , 0 0 0

x y

x y

x y

Converting inequality constraints into equality by adding slack variables:

1 2 3

1 2 3

1 2 3

1 2 3

0 8 1 0 1 0 0 1 1, 0 0 0

0 4 1 0 0 1 0 9 , 0 0 0

0 1 2 6 0 0 1 1 2 , 0 0 0

4 8 0 0 0 0

x y S S S

x y S S S

x y S S S

z x y S S S

Z X Y S1 S2 S3 value Row Operation

S1 0 8 10 1 0 0 11000

S2 0 4 10 0 1 0 9000

S3 0 12 6 0 0 1 12000

Z 1 -4 -8 0 0 0 0

The entering variable column for a maximization problem is always the one that has

the most negative value in the Z row. Here, column Y is the entering variable column.

Leaving variable or point row is S2 because 9000/10 < 11000/10<12000/6, hence Y will

replace S2. S1 0 8 10 1 0 0 11000

Y 0 0.4 1 0 0.1 0 900

2 2/ 1 0R R

S3 0 12 6 0 0 1 12000

Z 1 -4 -8 0 0 0 0

S1 0 4 0 1 -1 0 2000 1 1 2

( 1 0 )R R R

Y 0 0.4 1 0 0.1 0 900

S3 0 9.6 0 0 -0.6 1 6600 3 3 2

( 6 )R R R

Z 1 -0.8 0 0 0.8 0 7200 4 4 2

(8 )R R R

Here, column X is the entering variable column now. Leaving variable or point row is

S1 because 2000/4 <6600/9.6< 900/0.4, hence X will replace S1.

X 0 1 0 0.25 -0.25 0 500

1 1/ 4R R

Y 0 0.4 1 0 0.1 0 900

S3 0 9.6 0 0 -0.6 1 6600

Z 1 -0.8 0 0 0.8 0 7200


Z X Y S1 S2 S3 value Row Operation

X 0 1 0 0.25 -0.25 0 500

Y 0 0 1 -0.1 0 0 700 2 2 1

( 0 .4 )R R R

S3 0 0 0 -2.4 -3 1 1800

3 3 1( 9 .6 )R R R

Z 1 0 0 0.2 0.6 0 7600 4 4 1

(0 .8 )R R R

The optimum production plan is to produce 500 units of product X and 700 units of

product Y. The maximum contribution we can get =$7,600. Shadow price for Division A

and B are $0.2 and $0.6 respectively. Division C: slack- 1800 hours available or no

shadow price.


Question2: Partex Furniture manufactures office furniture and home furniture and each

of these products passes through a cutting process and assembly process. One office

furniture, which makes average contribution of $50, takes six hours cutting time and four

hours assembly time; while one home furniture makes an average contribution of $40,

and takes three hours cutting time and eight hours assembly time. There is a maximum of

36 cutting hours and 48 assembly hours available each week. Cutters are paid $10 per

hour and assembly workers $15 per hour. Find the optimal production plan (maximize

contribution) for the company including shadow prices for cutting hours and assembly

hours.

Objective function, maximize contribution, 5 0 4 0z x y

Constraints:

C u tt in g h o u rs , 6 3 3 6

A s s e m b ly h o u rs , 4 8 4 8

x y

x y

Drawing Graph:

x y

0 12

6 0

x y

0 6

12 0

x y

0 5

4 0

Assembly

Hours

Cutting Hours

Feasible

Region

A

B

C

X

Y

Contribution

Line




ruler away from the origin (in the case of a maximization problem) by keeping it parallel

to the iso-contribution line, the last corner it touches of the feasible region represents the

optimum solution.

The optimum corner is B, the intersection of cutting hours and assembly hours.

6 3 3 6

4 8 4 8

, 4

, 4

x y

x y

o r x

a n d y

The optimum production plan is to produce 4 units of product X and 4 units of product

Y.

The maximum contribution we can get =50x4+40x4=$360


The optimum corner B is found in the intersection of cutting hours and assembly hours.

Therefore cutting hours and assembly hours are fully utilized and are referred to as

critical constraints. We can calculate shadow prices for the critical constraints. There is

no slack or unutilized resources available.

Shadow price for cutting hours:

Adding one extra hour for cutting, while leaving the other critical constraint unchanged.

6 3 3 7

4 8 4 8

, 4 .2 2

, 3 .8 9

x y

x y

o r x

a n d y

The increased contribution =50x4.22+40x3.89—$360=$6.6

The increase of $6.6 is the shadow price or the premium that the company would be

willing to pay for each extra hour of cutting.

Similarly we can calculate shadow price for assembly hours as well.




Cell Name

Original Value

Final Value

$C$2 Max. Contribution z $0.00 $360.00

Variable Cells

Cell Name

Original Value

Final Value Integer

$A$2 x 0 4 Contin

$B$2 y 0 4 Contin

Constraints


$B$11 Non-neg 4 $B$11>=0

Not Binding 4

$B$5 Cutting Hours 36 $B$5<=36 Binding 0

$B$7 Assembly Hours 48 $B$7<=48 Binding 0

$B$9 Non-neg 4 $B$9>=0

Not Binding 4

Binding: Critical Constraints (resource or hours fully utilized)

Constraints



$B$11 Non-neg 4 0 0 4 1E+30

$B$5 Cutting Hours 36 6.666666667 36 36 18

$B$7 Assembly Hours 48 2.5 48 48 24

$B$9 Non-neg 4 0 0 4 1E+30

Shadow price for cutting hours=$6.67 and for Assembly hours=$2.5

Variable

Lower Objective

Upper Objective

Cell Name Value

Limit Result

Limit Result

$A$2 x 4

0 160

4 360

$B$2 y 4

0 200

4 360


Simplex Method


Constraints:

C u tt in g h o u rs , 6 3 3 6

A s s e m b ly h o u rs , 4 8 4 8

x y

x y


1 2

1 2

1 2

0 6 3 1 . 0 3 6

0 4 8 0 1 . 4 8

5 0 4 0 0 0 . 0

x y S S

x y S S

Z x y S S

Z X Y S1 S2 value Row Operation

S1 0 6 3 1 0 36

S2 0 4 8 0 1 48

Z 1 -50 -40 0 0 0


the most negative value in the Z row. Here, column X is the entering variable column.

Leaving variable or point row is S1 because 6/36 < 4/48, hence X will replace S1.

X 0 1 0.5 0.167 0 6 1 1

/ 6R R

S2 0 1 2 0 0.25 12 2 2

/ 4R R

Z 1 -50 -40 0 0 0

X 0 1 0.5 0.167 0 6

S2 0 0 1.5 -0.167 0.25 6 2 2 1

R R R

Z 1 0 -15 8.34 0 300

3 1 3(5 0 )R R R

Here, column Y is the entering variable column now.

Leaving variable or point row is S2 because 6/1.5 < 6/0.5, hence Y will replace S2.

X 0 1 0.5 0.167 0 6

Y 0 0 1 -0.111 0.167 4 2 2

/ 1 .5R R

Z 1 0 -15 8.34 0 300

X 0 1 0 0.222 -0.083 4 1 1 2

( 0 .5 )R R R

Y 0 0 1 -0.111 0.167 4

Z 1 0 0 6.67 2.5 360 3 2 3

(1 5 )R R R

The optimum production plan is to produce 4 units of product X and 4 units of product

Y. The maximum contribution we can get =$360. Shadow price for cutting hours and

assembly hours are $6.67 and $2.5 respectively.


Question3: Millennium Cycle manufactures mountain bike and racing bike and apart

from raw materials it requires a mix of skilled and semi-skilled labours. The supply of

skilled and semi-skilled labour is limited to 2000 hours/month and 2500 hours/month

respectively. One mountain bike needs 10 hours of skilled and 5 hours of semi-skilled

labour hours while one racing bike needs 10 hours of skilled and 25 hours of semi-skilled

labour hours. The company receives average contribution of $50 from Mountain bikes

and $40 from racing bikes while skilled labours are paid $15 per hour and semi-skilled

labours are paid $10 per hour. At present, maximum demand for mountain bike and

racing bike is expected to be 150 units/month and 80 units/month respectively. Find the

optimal production plan (maximize contribution) for the company including shadow

prices for skilled labours and semi-skilled labours.


Constraints:

S k ille d L a b o u r , 1 0 1 0 2 0 0 0

S e m i-s k i lle d L a b o u r , 5 2 5 2 5 0 0

M o u n ta in B ik e , 1 5 0

R a c in g B ik e , 8 0

N o n -n e g a tiv i ty , 0 ; 0

x y

x y

x

y

x y

Drawing Graph:

x y

0 200

200 0

x y

0 100

500 0

x y

0 50

40 0

Semi-skilled

Skilled

A

B C

X

Y

CL

D

E

x≤150

y≤80

Feasible Region=ABCDE




ruler away from the origin (in the case of a maximization problem) by keeping it parallel

to the iso-contribution line, the last corner it touches of the feasible region represents the

optimum solution.

The optimum corner (last corner) is D, the intersection of skilled labour and x≤150.

1 5 0

1 0 1 0 2 0 0 0

, 5 0

x

x y

a n d y

The optimum production plan is to produce 150 units of mountain bikes and 50 units of

racing bike.

The maximum contribution we can get =50x150+40x50=$9500


The optimum corner D is found in the intersection of skilled labour and x≤150. Therefore

skilled labours are fully utilized and are referred to as critical constraints. We can

calculate shadow price for the skilled labour. There is slack or unutilized hours available

for semi-skilled labours; hence no shadow price for semi-skilled labour.

Shadow price for skilled labour:

Adding one extra hour for skilled labour, while leaving the other constraint unchanged.

1 5 0

1 0 1 0 2 0 0 1

, 5 0 .1

x

x y

a n d y

The increased contribution =50x150+40x50.1—$9500=$4

The increase of $4 is the shadow price or the premium that the company would be willing

to pay for each extra hour of skilled labour.




Cell Name

Original Value Final Value

$C$2 Max. Contribution z $0.00 $9,500.00

Variable Cells

Cell Name


$A$2 x 0 150 Contin

$B$2 y 0 50 Contin

Constraints


$B$11 y 50 $B$11<=80

Not Binding 30

$B$14 Non-neg 150 $B$14>=0

Not Binding 150

$B$16 Non-neg 50 $B$16>=0

Not Binding 50

$B$5 Skilled Labours 2000 $B$5<=2000 Binding 0

$B$7 Semi-skilled Labours 2000 $B$7<=2500

Not Binding 500

$B$9 x 150 $B$9<=150 Binding 0

Binding: Critical Constraints (skilled labours are fully utilized),

Non-binding or non-critical constraints: 500 hours of slack or unutilized hours available

for semi-skilled labours.

Constraints



$B$11 y 50 0 80 1E+30 30

$B$14 Non-neg 150 0 0 150 1E+30

$B$16 Non-neg 50 0 0 50 1E+30

$B$5 Skilled Labours 2000 4 2000 200 500

$B$7 Semi-skilled Labours 2000 0 2500 1E+30 500

$B$9 x 150 10 150 50 25

Shadow price for skilled labours =$4

Cell Variable

Lower Objective

Upper Objective

Cell Name Value

Limit Result

Limit Result

$A$2 x 150

0 2,000

150 9,500

$B$2 y 50

0 7500

50 9500

When X=0, Y=50, Max contribution=$2000, and when Y=0, X=150, Max

contribution=$7500


Simplex Method


Constraints:

S k ille d L a b o u r , 1 0 1 0 2 0 0 0

S e m i-s k i lle d L a b o u r , 5 2 5 2 5 0 0

M o u n ta in B ik e , 1 5 0

R a c in g B ik e , 8 0


x y

x y

x

y

x y


1 2 3 4

1 2 3 4

1 2 3 4

1 2 3 4

1 2 3 4

0 1 0 1 0 1 0 0 0 2 , 0 0 0

0 5 2 5 0 1 0 0 2 , 5 0 0

0 1 0 0 0 1 0 1 5 0

0 0 1 0 0 0 1 8 0

5 0 4 0 0 0 0 0 0

x y S S S S

x y S S S S

x y S S S S

x y S S S S

z x y S S S S

Z X Y S1 S2 S3 S4 value Row Operation

S1 0 10 10 1 0 0 0 2000

S2 0 5 25 0 1 0 0 2500

S3 0 1 0 0 0 1 0 150

S4 0 0 1 0 0 0 1 80

Z 1 -50 -40 0 0 0 0 0

The entering variable column for a maximization problem is always the one that has the

most negative value in the Z row. Here, column X is the entering variable column.

Leaving variable or point row is S3 because 150/1<2000/10 < 2500/5, hence X will replace

S3.

S1 0 0 10 1 0 -10 0 500 1 1 3

( 1 0 )R R R

S2 0 0 25 0 1 -5 0 1750 2 2 3

( 5 )R R R

X 0 1 0 0 0 1 0 150

S4 0 0 1 0 0 0 1 80

Z 1 0 -40 0 0 50 0 7500 5 5 3

(5 0 )R R R

Here, column Y is the entering variable column now. Leaving variable or point row is S1

because 500/10 <1750/25< 80/1, hence Y will replace S1.

Y 0 0 1 0.1 0 -1 0 50

1 1/ 1 0R R

S2 0 0 25 0 1 -5 0 1750

X 0 1 0 0 0 1 0 150

S4 0 0 1 0 0 0 1 80

Z 1 0 -40 0 0 50 0 7500


Z X Y S1 S2 S3 S4 value Row Operation

Y 0 0 1 0.1 0 -1 0 50 1 1

/ 4R R

S2 0 0 0 -2.5 1 20 0 500 2 2 1

( 2 5 )R R R

X 0 1 0 0 0 1 0 150

S4 0 0 0 -0.1 0 1 1 30 4 4 1

R R R

Z 1 0 0 4 0 10 0 9500 5 5 1

( 4 0 )R R R

The optimum production plan is to produce 150 units of product X (Mountain Bike) and

50 units of product Y (Racing Bike). The maximum contribution we can get=$9,500.

Shadow price for skilled labour is $4 while semi-skilled labour has a slack of 500 hours

available, hence no shadow price.


Question4: ACI Agro can buy two types of fertilizer which contain the following

percentage of chemicals:

Nitrates Phosphates Potash

Type 1 18 5 2

Type 2 3 2 5

For a certain crop ACI need minimum 100 kg of Nitrates, 50 kg of Phosphates and 40 kg

of Potash. ACI currently buys 1000 kg of each type of fertilizer where type 1 costs $10/kg

and type 2 costs $5/kg. Find the optimal purchase plan (minimize cost) for the company.

Objective function, minimize cost, 1 0 5z x y

Constraints:

N itra te s , 0 .1 8 0 .0 3 1 0 0

P h o s p h a te s , 0 .0 5 0 .0 2 5 0

P o ta s h , 0 .0 2 0 .0 5 4 0


x y

x y

x y

x y

Drawing Graph:

x y

0 3333.34

556 0

x y

0 2500

1000 0

x y

0 800

2000 0

x y

0 4000

2000 0

Phosphates

Nitrates

A

B

C

X

Y

Iso-cost line

D

Feasible Region

Outside ABCD

Potash


The optimum point is found by drawing an example of an iso-cost line on the diagram

(any value of Z will do) and then placing a ruler against it. Then, by moving the ruler

towards the origin (in the case of a minimization problem) by keeping it parallel to the

iso-cost line, the last corner it touches of the feasible region represents the optimum

solution.

The optimum corner (last corner) is C, the intersection of potash and phosphates.

5 2 5 0 0 0

2 5 4 0 0 0

8 0 9 .5

, 4 7 6 .2

x y

x y

x

a n d y

The optimum purchase plan is to acquire 809.5 kg of type 1 and 476.2 kg of type 2

fertilizer.

The minimized cost we can get =10x809.5+5x476.2=$10,476

The current plan cost =10x1000+5x1000=$15,000

Cost saved=$15,000—$10,476=$4,524

Shadow price for Phosphates:

Reducing one extra kg requirement for Phosphates, while leaving the other constraint

unchanged.

5 2 4 9 0 0

2 5 4 0 0 0

7 8 5 .7 1

, 4 8 5 .7 1

x y

x y

x

a n d y

The minimized cost we can get =$10x785.71+$5x485.71=$10,285.69

Cost would be saved=$10,476—$10,285.69=$190.3

The reduction of $190.3 is the shadow price or the discount that the supplier would be

willing to give for each extra kg reduction in the requirement of Phosphates.



Variable Cells

Cell Name


$A$2 x 0 810 Contin

$B$2 y 0 476.1904762 Contin

Constraints


$B$11 Non-neg 809.5238095 $B$11>=0

Not Binding 809.5238095

$B$13 Non-neg 476.1904762 $B$13>=0

Not Binding 476.1904762

$B$5 Nitrates 160 $B$5>=100

Not Binding 60

$B$7 Phosphates 50 $B$7>=50 Binding 0

$B$9 Potash 40 $B$9>=40 Binding 0

Binding: Critical Constraints (we have exactly the minimum quantities required of

Phosphates and Potash)

Non-binding or non-critical constraints: We have 60 kg more than the minimum

quantities required of Nitrates.

Variable Cells

Final Reduced Objective Allowable Allowable

Cell Name Value Cost Coefficient Increase Decrease

$A$2 x 809.5238095 0 10 2.5 8

$B$2 y 476.1904762 0 5 20 1

Constraints



$B$11 Non-neg 809.5238095 0 0 809.5238095 1E+30

$B$13 Non-neg 476.1904762 0 0 476.1904762 1E+30

$B$5 Nitrates 160 0 100 60 1E+30

$B$7 Phosphates 50 190.4761905 50 50 15

$B$9 Potash 40 23.80952381 40 60 20


Simplex Method

Objective function, minimize cost, 1 0 5z x y

Constraints:

N itra te s , 0 .1 8 0 .0 3 1 0 0

P h o s p h a te s , 0 .0 5 0 .0 2 5 0

P o ta s h , 0 .0 2 0 .0 5 4 0


x y

x y

x y

x y

Surplus Variable Constraints

P1 0 .1 8 0 .0 3x y 100

P2 0 .0 5 0 .0 2x y 5 0

P3 0 .0 2 0 .0 5x y 4 0

Min 1 0 5x y Z

Now converting the minimization problem into dual problem (maximization problem):

1 2 3

1 2 3

0 .1 8 0 .0 5 0 .0 2 1 0

0 .0 3 0 .0 2 0 .0 5 5

P P P

P P P

Objective function, maximize, 1 2 31 0 0 5 0 4 0P P P z


1 2 3 1 2

1 2 3 1 2

1 2 3 1 2

0 + 0 .1 8 0 .0 5 0 .0 2 1 0 1 0

0 + 0 .0 3 0 .0 2 0 .0 5 0 1 5

1 0 0 5 0 4 0 0 0 0

P P P S S

P P P S S

z P P P S S

Z P1 P2 P3 S1 S2 value Row Operation

S1 0 0.18 0.05 0.02 1 0 10

S2 0 0.03 0.02 0.05 0 1 5

Z 1 -100 -50 -40 0 0 0


the most negative value in the Z row. Here, column P1 is the entering variable column.

Leaving variable or point row is S1, hence P1 will replace S1.

P1 0 1 0.2778 0.111 5.556 0 55.556 1 1

/ 0 .1 8R R

S2 0 0.03 0.02 0.05 0 1 5

Z 1 -100 -50 -40 0 0 0



P1 0 1 0.2778 0.111 5.556 0 55.556

S2 0 0 0.0116 0.046 -0.166 1 3.3333 2 2 1

( 0 .0 3)R R R

Z 1 0 -22.22 -28.88 555.55 0 5555.6

3 3 1(1 0 0 )R R R

Here, column P3 is the entering variable column now. Leaving variable or point row is

S2, hence P3 will replace S2.

P1 0 1 0.2778 0.111 5.556 0 55.556

P3 0 0 0.252 1 -3.6 21.74 72.46 2 2

/ 0 .0 4 6R R

Z 1 0 -22.22 -28.88 555.55 0 5555.6

P1 0 1 0.25 0 5.955 -2.41 47.51 1 1 2

( 0 .1 1 1)R R R

P3 0 0 0.252 1 -3.6 21.74 72.46

Z 1 0 -14.94 0 451.58 627.85 7648.24 3 3 2

( 2 8 .8 8 )R R R

Here, column P2 is the entering variable column now. Leaving variable or point row is

P1, hence P2 will replace P1.

P2 0 4 1 0 23.82 -9.64 190.04 1 1

/ 0 .2 5R R

P3 0 0 0.252 1 -3.6 21.74 72.46

Z 1 0 -14.94 0 451.58 627.85 7648.24


P2 0 4 1 0 23.82 -9.64 190.04

P3 0 -1 0 1 -9,6 24.17 24.57 2 2 1( 0 .2 5 2 )R R R

Z 1 59.76 0 0 807.45 483.82 10487.43 3 3 1(1 4 .9 4 )R R R

The value of S1=X=807 units, and the value of S2= Y=483 units and the minimized

cost=$10,487.43


Question5: Pure Gasoline Company operates two refineries with different production

capacities. Refinery A can produce 4,000 gallons per day of Super Unleaded gasoline,

2,000 gallons per day of Regular Unleaded gasoline, and 1,000 per day of Diesel fuel. On

the other hand Refinery B can produce 1,000 gallons per day of Super Unleaded, 3,000

gallons per day of Regular Unleaded, and 4,000 gallons per day of Diesel fuel. The

company has made a contract with an automobile manufacturer to provide 24,000 gallons

of Super Unleaded, 42,000 gallons of Regular Unleaded, and 36,000 gallons of Diesel fuel.

Determine 1) the number of days the company should operate each refinery in order to

meet the terms of the contract most economically, 2) the minimum cost, and 3) what

grade(s) of fuel would be overproduced if the cost of running Refinery A is $1,500 per day

and Refinery B is $2,400 per day. Objective function, minimize cost, 1 5 0 0 2 4 0 0z x y

Constraints:

S u p e r U n le a d e d , 4 0 0 0 1 0 0 0 2 4 , 0 0 0

R e g u la r U n le a d e d , 2 0 0 0 3 0 0 0 4 2 , 0 0 0

D ie s e l, 1 0 0 0 4 0 0 0 3 6 , 0 0 0


x y

x y

x y

x y

Surplus Variable Constraints

P1 4 0 0 0 1 0 0 0x y 2 4 , 0 0 0

P2 2 0 0 0 3 0 0 0x y 4 2 , 0 0 0

P3 1 0 0 0 4 0 0 0x y 3 6 , 0 0 0

Min 1 5 0 0 2 4 0 0x y Z

Now converting the minimization problem into dual problem (maximization problem):

1 2 3

1 2 3

4 0 0 0 2 0 0 0 1 0 0 0 1 5 0 0

1 0 0 0 3 0 0 0 4 0 0 0 2 4 0 0

P P P

P P P

Objective function, maximize, 1 2 32 4 , 0 0 0 4 2 0 0 0 3 6 0 0 0P P P z


1 2 3 1 2

1 2 3 1 2

1 2 3 1 2

0 + 4 0 0 0 2 0 0 0 1 0 0 0 1 0 1 5 0 0

0 + 1 0 0 0 3 0 0 0 4 0 0 0 0 1 2 4 0 0

2 4 , 0 0 0 4 2 0 0 0 3 6 0 0 0 0 0 0

P P P S S

P P P S S

z P P P S S


S1 0 4000 2000 1000 1 0 1500

S2 0 1000 3000 4000 0 1 2400

Z 1 -24000 -42000 -36000 0 0 0


the most negative value in the Z row. Here, column P2 is the entering variable column.

Leaving variable or point row is S1, hence P2 will replace S1.



P2 0 2 1 0.5 0.0005 0 0.75 1 1

/ 2 0 0 0R R

S2 0 1000 3000 4000 0 1 2400

Z 1 -24000 -42000 -36000 0 0 0

P2 0 2 1 0.5 0.0005 0 0.75

S2 0 -5000 0 2500 -1.5 1 150 2 2 1( 3 0 0 0 )R R R

Z 1 60000 0 -15000 21 0 31500 3 3 1( 4 2 0 0 0 )R R R

Here, column P3 is the entering variable column. Leaving variable or point row is S2,

hence P3 will replace S2.

P2 0 2 1 0.5 0.0005 0 0.75

P3 0 -2 0 1 -0.0006 0.0004 0.06 2 2/ 2 5 0 0R R

Z 1 60000 0 -15000 21 0 31500

P2 0 3 1 0 0.0008 -0.0002 0.72 1 1 2( 0 .5 )R R R

P3 0 -2 0 1 -0.0006 0.0004 0.06

Z 1 30000 0 0 12 6 32400 3 3 2(1 5 0 0 0 )R R R

The value of S1=X=12 days, and the value of S2= Y=6 days and the minimized

cost=$32,400. The company should operate Refinery A for 12 days and Refinery B for 6

days to fulfil the contract with associated minimum cost of $32,400.


Question6: In a chair manufacturing company Lounge chairs and Swivel chairs are made

using the equipment of two departments: I and II. It requires one hour in each

department to make a lounge chair, but making a swivel chair takes one hour in

department I and two hours in department II. Department I has four hours of time

available, and II has six hours available. Each Lounge chair made and sold contributes $1

to profit, and each Swivel chair contributes $0.50 to profit. Find the optimal production

plan (maximize contribution) for the company.


Cell Name Original Value Final Value

$A$11 Objective 0 4

Variable Cells

Cell Name Original Value Final Value Integer

$A$2 x 0 4 Contin

$B$2 y 0 0 Contin

Constraints


$A$7 Dept1 4 $A$7<=4 Binding 0

$A$9 Dept 2 4 $A$9<=6 Not Binding 2

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6

Dept 1 Dept 2 objective


Constraints



$A$7 Dept1 4 1 4 2 4

$A$9 Dept 2 4 0 6 1E+30 2

Question7: Ace Rubber Company manufactures two types of tires: Model P (the

premium) and Model R (the regular). Model P sells for $95 per tire and costs $85 per tire

to make, whereas Model R sells for $50 per tire and costs $42 per tire to make. To make

one model P tire, it requires two hours on machine A and four hours on machine B. On

the other hand, to make one model R tire, it takes 9 hours on machine A and three hours

on machine B. Production scheduling indicates that during the coming week Machine A

will be available for at most 36 hours and machine B for at most 42 hours. How many of

each tire should the company make in the coming week in order to maximize its profit?

What is this maximum profit?



$A$11 Objective 0 106

Variable Cells


$A$2 x 0 9 Contin

$B$2 y 0 2 Contin

Constraints


$A$7 MachineA 36 $A$7<=36 Binding 0

$A$9 Machine B 42 $A$9<=42 Binding 0

$A$2 x 9 $A$2>=0 Not Binding 9

$B$2 y 2 $B$2>=0 Not Binding 2

Constraints



$A$7 MachineA 36 0.066666667 36 90 15

$A$9 Machine B 42 2.466666667 42 30 30


Question8: Star Insulating Company manufactures two types of storm windows: Model

H (the heavy duty) and Model R (the regular). Model H sells for $45 per window and cost

$36 per window to make, whereas Model R sells for $35 per window and costs $27 per

window to make. To make one model H window, it requires 4 hours on machine A and 3

hours on machine B. On the other hand, to make one model R window, it takes 5 hours

on machine A and 2 hours on machine B. Production scheduling indicates that during the

coming week Machine A will be available for at most 30 hours and machine B for at most

19 hours. How many of each window should the company make in the coming week in

order to maximize its profit? What is this maximum profit?

0

2

4

6

8

10

12

14

16

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

Machine A Machine B Objective




$A$11 Objective 0 66

Variable Cells


$A$2 x 0 5 Contin

$B$2 y 0 2 Contin

Constraints


$A$7 MachineA 30 $A$7<=30 Binding 0




Constraints



$A$7 MachineA 30 0.571428571 30 17.5 4.666666667

$A$9 Machine B 19 2.571428571 19 3.5 7

Question9: Safety Lock Company makes two kinds of locks: Model SS (Super safe) and

Model S (safe). Each model SS lock sells for $24 and costs $19 to make, while each model

S lock sells for $18 and costs $15 to make. Model SS requires 3 hours on machine A and 2

hours on machine B, whereas model S requires 7 hours on machine A and 1 hour on

machine B. During the coming week machine A will be available for no more than 42

0

1

2

3

4

5

6

7

8

9

10

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5



hours and machine B for no more than 17 hours. Determine the number of each kind of

lock to be made in the coming week for the company to maximize its profit. What is this

maximum profit?



$A$9 Obj 0 44

Variable Cells


$A$2 x 0 7 Contin

$B$2 y 0 3 Contin

Constraints


$A$5 Machine A 42 $A$5<=42 Binding 0




Constraints



$A$5 Machine A 42 0.090909091 42 77 16.5

$A$7 Machine B 17 2.363636364 17 11 11

0

2

4

6

8

10

12

14

16

18

0

0.5 1

1.5 2

2.5 3

3.5 4

4.5 5

5.5 6

6.5 7

7.5 8

8.5 9

9.5 10

10

.5 11

11

.5 12

12

.5 13

13

.5 14



Question10: XYZ Steel Company manufactures two kinds of wrought-iron rails: Model E

(the elegant) and Model D (the distinctive). Model E rails sell for $59 and cost $50 to

make, whereas Model D rails sell for $48 and cost $41 to make. To make one model E rail

requires 2 hours on machine A, 1 hour on machine B and 4 hours on machine C. On the

other hand, to make one model D rail requires 1 hour on machine A, 2 hours on B, and 5

hours on C. Production scheduling indicates that, during the coming week machine A will

be available for at most 30 hours, machine B for at most 24 hours, and machine C for at

most 72 hours. Find the number of each kind of rail to be made in the coming week in

order for the company to maximize its profit. What is this maximum profit? At the

maximum, which machines, if any are not fully utilized?

Question11: A special food for athletes is to be developed from two foods: Food X and

Food Y. The new food is to contain at least 16mg of vitamin A, at least 20 mg of vitamin

B, and at least 12mg of vitamin C. Each pound of food X costs $1.5 and contains 1mg of

vitamin A, 5mg of vitamin B and 1mg of vitamin C. On the other hand each pound of

food Y costs $2.5 and contains 2mg of vitamin A, 1mg of vitamin B and 1mg of vitamin C.

How many pounds of each food should be used in the mixture in order to meet the

preceding requirements at a minimum cost? What is this minimum cost?

Question12: Repeat problem 11 if, in addition, the amount of food Y in the mixture must

be no more than one and half the amount of food X.

Objective Cell (Min)


$A$12 Obj 0 22

Variable Cells


$A$2 x 0 8 Contin

$B$2 y 0 4 Contin

Constraints


$A$10 Mixture 16 $A$10>=0 Not Binding 16

$A$4 Vit A 16 $A$4>=16 Binding 0

$A$6 Vit B 44 $A$6>=20 Not Binding 24

$A$8 Vit C 12 $A$8>=12 Binding 0



Constraints



$A$10 Mixture 16 0 0 16 1E+30

$A$4 Vit A 16 1 16 3.2 4

$A$6 Vit B 44 0 20 24 1E+30

$A$8 Vit C 12 0.5 12 4 2


Question13: Strong Steel Company operates two steel mills with different production

capacities. Mill I can produce 1,000 tons per day of AAA steel, 3,000 tons per day of AA

steel, and 5,000 tons per day of A steel. Mill II can produce 2,000 tons per day of each

grade of steel. The company has made a contract with the construction firm to provide

24,000 tons of AAA steel, 32,000 tons of AA steel, and 40,000 tons of A steel. The cost of

running mill I and II are $1,400 per day and $1,000 per day. Find the optimal production

plan.

0

5

10

15

20

25

0

0.5 1

1.5 2

2.5 3

3.5 4

4.5 5

5.5 6

6.5 7

7.5 8

8.5 9

9.5 10

10

.5 11

11

.5 12

12

.5 13

13

.5 14

14

.5 15

15

.5 16

Vit A Vit B Vit C Mixture Obj


0

5

10

15

20

25 AAA Steel AA Steel A Steel Objective




$A$13 Obj 0 15600

Variable Cells


$A$2 x 0 4 Contin

$B$2 y 0 10 Contin

Constraints


$A$10 A Steel 40000 $A$10>=40000 Binding 0

$A$4 AAA Steel 24000 $A$4>=24000 Binding 0

$A$7 AA Steel 32000 $A$7>=32000 Binding 0

$A$2 x 4 $A$2>=0

Not Binding 4

$B$2 y 10 $B$2>=0

Not Binding 10

Constraints



$A$10 A Steel 40000 0 40000 -1.81899E-12 1E+30

$A$4 AAA Steel 24000 0.05 24000 0 13333.33333

$A$7 AA Steel 32000 0.45 32000 40000 0

Question14: ABC dairy company wishes to make a new cheese from two of its current

cheeses: Cheese X and Cheese Y. The mixture is to weigh to more than 4 pounds and is to

contain at least 6 ounces of the sharpness ingredient S. Each pound of X costs $4 and

contains 3 ounces of S, whereas each pound of Y costs $1 and contain 1 ounce of S. How

many pounds of each cheese should be used in the mixture in order to meet these

requirements at a minimum cost? What is this minimum cost?




$A$13 Obj 0 6

Variable Cells


$A$2 x 0 0 Contin

$B$2 y 0 6 Contin

Constraints


$A$4 Mixture 6 $A$4>=4 Not Binding 2

$A$7 Ingredient S 6 $A$7>=6 Binding 0

$A$2 x 0 $A$2>=0 Binding 0


Question15: A wholesaler has 9,600 feet of space available, and $5,000 to buy

merchandise of types A, B and C. Type A costs $4 per unit and requires 4 feet of storage

space in the warehouse. B costs $10 per unit and requires 8 feet of space. C costs $5 per

unit and requires 6 feet of space. Only 500 units of type A are available to the wholesaler.

Assuming that the wholesaler expects to make a profit of $1 on each unit of A bought and

stocked, $3 per unit on B, and $2 per unit on C, how many units of each should be bought

and stocked in order to maximize his profit, and what is this maximum profit?

Question16: A merchant plans to sell two models of home computers at costs of $250 and

$400, respectively. The $250 model yields a profit of $45 and the $400 model yields a

profit of $50. The merchant estimates that the total monthly demand will not exceed 250

units. Find the number of units of each model that should be stocked in order to

maximize profit. Assume that the merchant does not want to invest more than $70,000 in

computer inventory.

0

2

4

6

8

10

12

14

16

18

0 0.5 1 1.5 2 2.5 3 3.5 4

Mixture Ingredient S Obj


Question17: A fruit grower has 150 acres of land available to raise two crops, A and B. It

takes one day to trim an acre of crop A and two days to trim an acre of crop B, and there

are 240 days per year available for trimming. It takes 0.3 day to pick an acre of crop A

and 0.1 day to pick an acre of crop B, and there are 30 days per year available for

picking. Find the number of acres of each fruit that should be planted to maximize profit,

assuming that the profit is $140 per acre for crop A and $235 per acre for B.

Question18: A grower has 50 acres of land for which she plans to raise three crops. It

costs $200 to produce an acre of carrots and the profit is $60 per acre. It costs $80 to

produce an acre of celery and the profit is $20 per acre. Finally, it costs $140 to produce

an acre of lettuce and the profit is $30 per acre. Use the simplex method to find the

number of acres of each crop she should plant in order to maximize her profit. Assume

that her cost cannot exceed $10,000.

Question19: A fruit juice company makes two special drinks by blending apple and

pineapple juices. The first drink uses 30% apple juice and 70% pineapple, while the

second drink uses 60% apple and 40% pineapple. There are 1000 liters of apple and 1500

liters of pineapple juice available. If the profit for the first drink is $0.60 per liter and

that for the second drink is $0.50, use the simplex method to find the number of liters of

each drink that should be produced in order to maximize the profit.

Question20: A manufacturer produces three models of bicycles. The time (in hours)

required for assembling, painting, and packaging each model is as follows.

Model A Model B Model C

Assembling 2 2.5 3

Painting 1.5 2 1

Packaging 1 0.75 1.25

The total time available for assembling, painting, and packaging is 4006 hours, 2495

hours and 1500 hours, respectively. The profit per unit for each model is $45 (Model A),

$50 (Model B), and $55 (Model C). How many of each type should be produced to obtain

a maximum profit?

Question21: Suppose in Exercise 25 the total time available for assembling, painting, and

packaging is 4000 hours, 2500 hours, and 1500 hours, respectively, and that the profit per

unit is $48 (Model A), $50 (Model B), and $52 (Model C). How many of each type should

be produced to obtain a maximum profit?

Question22: A company has budgeted a maximum of $600,000 for advertising a certain

product nationally. Each minute of television time costs $60,000 and each one-page

newspaper ad costs $15,000. Each television ad is expected to be viewed by 15 million

viewers, and each newspaper ad is expected to be seen by 3 million readers. The

company’s market research department advises the company to use at most 90% of the

advertising budget on television ads. How should the advertising budget be allocated to

maximize the total audience?

Question23: Rework Exercise 27 assuming that each one-page newspaper ad costs

$30,000.


Question24: An investor has up to $250,000 to invest in three types of investments. Type

A pays 8% annually and has a risk factor of 0. Type B pays 10% annually and has a risk

factor of 0.06. Type C pays 14% annually and has a risk factor of 0.10. To have a well-

balanced portfolio, the investor imposes the following conditions. The average risk factor

should be no greater than 0.05. Moreover, at least one-fourth of the total portfolio is to be

allocated to Type A investments and at least one-fourth of the portfolio is to be allocated

to Type B investments. How much should be allocated to each type of investment to

obtain a maximum return?

Question25: An investor has up to $450,000 to invest in three types of investments. Type

A pays 6% annually and has a risk factor of 0. Type B pays 10% annually and has a risk

factor of 0.06. Type C pays 12% annually and has a risk factor of 0.08. To have a well-

balanced portfolio, the investor imposes the following conditions. The average risk factor

should be no greater than 0.05. Moreover, at least one-half of the total portfolio is to be

allocated to Type A investments and at least one-fourth of the portfolio is to be allocated

to Type B investments. How much should be allocated to each type of investment to

obtain a maximum return?

Question26: An accounting firm has 900 hours of staff time and 100 hours of reviewing

time available each week. The firm charges $2000 for an audit and $300 for a tax return.

Each audit requires 100 hours of staff time and 10 hours of review time, and each tax

return requires 12.5 hours of staff time and 2.5 hours of review time. What number of

audits and tax returns will bring in maximum revenue?

Question27: A small petroleum company owns two refineries. Refinery 1 costs $20,000

per day to operate, and it can produce 400 barrels of high-grade oil, 300 barrels of

medium-grade oil, and 200 barrels of low-grade oil each day. Refinery 2 is newer and

more modern. It costs $25,000 per day to operate, and it can produce 300 barrels of high-

grade oil, 400 barrels of medium-grade oil, and 500 barrels of low-grade oil each day. The

company has orders totaling 25,000 barrels of high-grade oil, 27,000 barrels of medium-

grade oil, and 30,000 barrels of low-grade oil. How many days should it run each refinery

to minimize its costs and still refine enough oil to meet its orders?







grade oil, and 30,000 barrels of low-grade oil. How many days should it run each refinery

to minimize its costs and still refine enough oil to meet its orders?

Question29: Two dietary drinks are used to supply protein and carbohydrates. The first

drink provides 1 unit of protein and 3 units of carbohydrates in each litre. The second

drink supplies 2 units of protein and 2 units of carbohydrates in each litre. An athlete

requires 3 units of protein and 5 units of carbohydrates. The first drink costs $4 per litre

and the second costs $2 per litre. Find the amount of each drink the athlete should

consume to minimize the cost and still meet the minimum dietary requirements.


Question30: An athlete uses two dietary drinks that provide the nutritional elements

listed in the following table.

Drink Protein Carbohydrates Vitamin D

I 4 2 1

II 1 5 1

Drink I costs $7 per litre and drink II costs $4 per litre. Find the combination of drinks of

minimum cost that will meet the minimum requirements of 4 units of protein, 10 units of

carbohydrates, and 3 units of vitamin D.

Question31: A company has three production plants, each of which produces three

different models of a particular product. The daily capacities (in thousands of units) of

the three plants are as follows.

Model 1 Model 2 Model 3

Plant 1 8 4 8

Plant 2 6 6 3

Plant 3 12 4 8

The total demand for Model 1 is 300,000 units, for Model 2 is 172,000 units, and for

Model 3 is 249,500 units. Moreover, the daily operating cost for Plant 1 is $55,000, for

Plant 2 is $60,000, and for Plant 3 is $60,000. How many days should each plant be

operated in order to fill the total demand, and keep the operating cost at a minimum?

Question32: The Company in Exercise 29 has lowered the daily operating cost for Plant 3

to $50,000. How many days should each plant be operated in order to fill the total

demand, and keep the operating cost at a minimum?







grade oil, and 40,000 barrels of low-grade oil. How many days should the company run

each refinery to minimize its costs and still meet its orders?

Question34: A steel company has two mills. Mill 1 costs $70,000 per day to operate, and it

can produce 400 tons of high-grade steel, 500 tons of medium-grade steel, and 450 tons of

low-grade steel each day. Mill 2 costs $60,000 per day to operate, and it can produce 350

tons of high-grade steel, 600 tons of medium-grade steel, and 400 tons of low-grade steel

each day. The company has orders totaling 100,000 tons of high-grade steel, 150,000 tons

of medium-grade steel, and 124,500 tons of low-grade steel. How many days should the

company run each mill to minimize its costs and still fill the orders?

Date post:	05-Mar-2021
Category:	Documents
Upload:	others
View:	0 times
Download:	0 times

Notes Prepared By: Mohammad Kamrul...

Documents