Maxima and Minima Problems

Absolute Extrema

In such a case, the number is called the absolute maximum

value of f on I.

Definition. A function f has an absolute maximum value on an interval I if there is some number c in I such that

for each x in I. xfcf

f c

In such a case, the number is called the absolute minimum

value of f on I.

Definition. A function f has an absolute minimum value on an interval I if there is some number c in I such that

for each x in I. f c f x

f c

Definition. An absolute extremum of a function f on an interval I is either an absolute maximum value or an absolute minimum value of f on I.

Illustration: Determine the absolute extrema of the functions represented by the following graphs.

The absolute minimum of the function occurs at b.

The absolute maximum of the function occurs at a.

The absolute minimum of the function occurs at a.

The absolute maximum of the function occurs at b.

Figure 4.4.1(d)

The absolute maximum of the function occurs at c.

The absolute minimum of the function occurs at a.

The absolute minimum of the function occurs at c.

The absolute maximum of the function occurs at b.

Theorem The Extreme-Value Theorem

If a function f is continuous on a closed interval [a,b] then f has an absolute maximum value and an absolute minimum value on [a,b].

A

B

C

D

E

F

G

Relative minimum pts:

C, E

Absolute minimum pt:

G

Relative maximum pts:

B, D,

Absolute maximum pt:

DF

1. Find the critical number(s) of f.

2. Find the value of f at each of the critical numbers of f on .

3. Find the values of and .

4. The largest of the values from steps 2 and 3 is the absolute maximum value of f; the smallest of the values is the absolute minimum value of f.

How do you find the absolute extrema of a continuous function f on the closed interval [a, b]?

b,a af bf

if .Find the absolute extrema of on

Example1

0513 xx

05143' 2 xxxf

1 or 53x x

f 2,0 xxxxf 57 23

Solution:

• Solve for the critical numbers of f on (0,2).

Does f have absolute maximum and minimum values?

Yes

Candidates for absolute extrema:

3 27 5 , 0,2f x x x x x

262 f

1 23

3 27f

2x

0x

3

1x

xf

00 f

Conclusion

f has an absolute minimum value of -23/27 at x = 1/3.

f has an absolute maximum value of 26 at x=2

3 27 5 , 0,2f x x x x x

(2,26)

(0.33,-0.85)

if .Find the absolute extrema of g on [-3,-1]

Example 2

2' 3 5 0g x x

3 5 4g x x x

Solution:

• Solve for the critical numbers of g on (-3,-1).

Does f have absolute maximum and minimum values? Yes

For what value/s of x will g’(x) be zero?

Thus, g has no critical number on [-3,-1].

None

Compute the function values of g at the endpoints.

3 5 4, 3, 1g x x x x

31 1 5 1 4

10

g

1x

3x 33 3 5 3 4

46

g

g has an absolute minimum value of -46 at x = -3.

g has an absolute maximum value of -10 at x = -1

Find the absolute extrema of the following functions on the indicated interval.

Exercise

2 on ( 2,3]f x x

4 on [2,5)f x xa.

b.

c.

d. on [ 1,2]2

xf x

x

3 on [ 3, )f x x

Steps in Solving Max-Min Problems

Step 1. Understand the problem.

Read the problem carefully. Identify the information you need to solve the problem. What is unknown? What is given? What is required?

Step 2. Develop a mathematical model of the problem.

Draw pictures and label the parts that are important to the problem. Introduce a variable to represent the quantity to be maximized or minimized. Write a function that relates the variable/s to the problem.

Step 3. Find the domain of the function.Determine what values of the variable

make sense in the problem.

Step 4. Identify the critical points and endpoints.

Step 5. Solve the mathematical model.

Step 6. Interpret the solution.

A piece of wire 10 feet long is cut into two pieces. One piece is bent into the shape of a circle and the other into the shape of the square. How should the wire be cut so that the combined area of the two figures is as small as possible?

Example 1

x x10

Total Area =Area of Circle + Area of Square

Total Length of Wire

= Circumference of Circle + Perimeter of Square

Solution:

4

10 xs

xs 104

2

xr

2

2

x

xr 2

2

4

10

x

r s

Total Area =

2

2

x2

4

10

x

Total Area =

16

20100

4

22 xxxxfA

where 0,10x

5' ' 0

2 4 8

x xA f x

10

4x

.4

10

44

x

10x

0x

2510 7.96f

2

10 125 100

4 4

3.5

f

250 6.25

4f

2

( )2

xf x

2

4

10

x

Total Area: , 0,10x

Absolute minimum

Absolute minimum value of f at x = 4.4 is 3.5

Thus, 4.4 ft and 5.6 ft of wire should be used to form the circle and the square, respectively, to have the smallest combined area of f(4.4) = 3.5 sq. ft.

x = 4.4

10 - x = 5.6

Theorem

Suppose a function f is continuous on an interval I containing the number c. If f(c) is a relative extremum of f on I and c is the only number in I for which f has a relative extremum, then f(c) is an absolute extremum of f on I.

Example 2

A farmer has 2400 feet of fencing materials and wants to fence off a rectangular field that borders a straight river. He needs no fence along the river. What are the dimensions of the field that has the largest area?

w

l

Let l be the length (in feet) of the

field parallel to the river

w be the width (in feet) of the

rectangular field

Solution:

If A is the area of the

rectangular field, then

wwwfA 22400

600w

lwA

24002 wl (length of fencing materials )

042400' wwf

22400 2 , 0,1200f w w w w

4'' wf and '' 600 4 0f Thus, f has a relative maximum at w = 600.

w

l

2400 2l w (length in terms of width )

Since the only relative maximum value of f is at w = 600, then the absolute maximum value of f occurs when w = 600.

Therefore, using 2400 feet of fencing materials the rectangular field must be 1200 ft by 600 ft to have the largest possible area of f(600) = 720,000 sq.ft.

w = 600

l = 2400 – 2w = 2400 – 2 (600)

= 1200

Example 3Find two numbers whose difference is

100 and whose product is a minimum.

100P f x x x 100x-yxyP

Solution:

Let x and y be the two numbers

P be the product of the two numbers

where 100y x

2 100 , ,f x x x x

2'' xf and '' 50 2 0f ' 2 100 0f x x

Hence, f has a relative minimum value at x = 50.

Since the only relative minimum of f is when x = 50, then f(50) is the absolute minimum value.

Therefore, the two numbers in the problem are 50 and - 50. The minimum product is f(50) = - 2500.

2 100 , ,f x x x x

50x

x = 50 y = x - 100 = - 50

Example 4Find the height of the right circular cylinder of maximum

volume which can be inscribed in a sphere of radius 9 cm.

2

814

hh

22 29

2

hr

2V r h

Solution:

Let h be the height (in cm) of the cylinder

22 81

4

hr

V(h) be the volume (in sq.cm) of the cylinder

h/29

r

r be the radius (in cm) of the cylinder

3181 , 0,18

4V h h h h

0 0V

3 408 .91 12V

23' 81 0

4V h h

V has an absolute maximum at h = 18.4

Therefore, the height of the right circular cone which can be inscribed in the sphere of radius 9 cm should be 18.4 cm to have a maximum volume of 3124.9 cm2.

108 18.4h

3181 , 0,18

4V h h h h

18 0V