Home > Documents > 03 Relative and Absolute Extrema - Handout

# 03 Relative and Absolute Extrema - Handout

Date post: 15-Jan-2016
Category:
View: 28 times
Description:
Math 55 chapter 3
15
Relative and Absolute Extrema Math 55 - Elementray Analysis III Institute of Mathematics University of the Philippines Diliman Math 55 Relative and Absolute Extrema 1/ 15
Transcript

Relative and Absolute Extrema

Math 55 - Elementray Analysis III

Institute of Mathematics

University of the Philippines

Diliman

Math 55 Relative and Absolute Extrema 1/ 15

Recall

For a function f(x)

a is a critical number of f if either f 0(a) = 0 or f 0(a) doesnot exist.

The relative extrema of f occurs at the critical number, i.e.if a has a relative extremum at a, then a is a criticalnumber of f .

If f is continuous on a closed interval [a, b], then f attainsits abolute extrema on [a, b] at the critical points in [a, b] orat the endpoints a and b.

Math 55 Relative and Absolute Extrema 2/ 15

Relative and Absolute Extrema

Definition

A function f(x, y) has a

relative minimum at (a, b) if f(a, b) f(x, y);

relative maximum at (a, b) if f(a, b) � f(x, y)

for all (x, y) on some disk centered at (a, b). If the aboveinequality holds for all (x, y) in the domain of f , then f has anabsolute minimum or absolute maximum at (a, b).

Math 55 Relative and Absolute Extrema 3/ 15

Critical Point

Definition

The point (a, b) is a critical point or stationary point off(x, y) if either

rf(a, b) = ~0 (i.e., fx

(a, b) = 0 and f

y

(a, b) = 0),

one of fx

(a, b) or fy

(a, b) does not exist.

Math 55 Relative and Absolute Extrema 4/ 15

Relative Extrema

Theorem

If f(x, y) has a relative extremum at (a, b), then (a, b) is a

critical point of f(x, y) and in fact, rf(a, b) = ~0.

Proof. Define g(x) = f(x, b) and suppose f has a relativeextremum at (a, b). Then g(x) has relative extremum (of thesame kind) at x = a, i.e., g0(a) = 0. Thus, g0(a) = f

x

(a, b) = 0.

By defining h(y) = f(a, y) and using similar aruments, we havef

y

(a, b) = 0.

This means rf(a, b) = ~0 and hence (a, b) is a critical point of f⌅

Math 55 Relative and Absolute Extrema 5/ 15

Consider f(x, y) = x

2 � y

2. Note that

f

x

(x, y) = 2x f

y

(x, y) = �2y

so the only critical point of f is (0, 0).

Graph of f(x, y) = x

2 � y

2

In this case we call (0, 0) a saddle point of f .

Math 55 Relative and Absolute Extrema 6/ 15

Second Derivatives Test

Suppose (a, b) is a critical point of f(x, y), with second orderpartial derivatives of f continuous on some region containing(a, b). Define D = f

xx

(a, b)fyy

(a, b)� [fxy

(a, b)]2. Then f has

a relative minimum at (a, b) if D > 0 and f

xx

> 0;

a relative maximum at (a, b) if D > 0 and f

xx

< 0;

a saddle point at (a, b) if D < 0.

Remarks.

No conclusion can be made if D = 0.

The formula for D can be written as the determinant of the

matrix

f

xx

f

xy

f

yx

f

yy

�.

Math 55 Relative and Absolute Extrema 7/ 15

Example

Find and classify all critical points of f(x, y) = 4+x

3+ y

3� 3xy

Solution. Solving the system

(f

x

(x, y) = 0

f

y

(x, y) = 0)

(3x2 � 3y = 0

3y2 � 3x = 0

gives the solutions (0, 0) and (1, 1).

Note that fxx

(x, y) = 6x; f

yy

(x, y) = 6y; f

xy

(x, y) = �3

f

xx

f

yy

f

xy

D

(0, 0) 0 0 �3 0� (�3)2 = �9(1, 1) 6 6 �3 6(6)� (�3)2 = 27

Hence f(x, y) has a saddle point at (0, 0) (sinceD < 0) and a relative minimum at (1, 1) (sinceD > 0 and f

xx

> 0)Graph of

f(x, y) = 4+x

3+y

3�3xy

Math 55 Relative and Absolute Extrema 8/ 15

Example

Consider the function f(x, y) = x

2

ye

�x

2�y

2

.To find the critical points, we solve the system

(f

x

(x, y) = �2xy(x2 � 1)e�x

2�y

2

= 0

f

y

(x, y) = x

2(1� 2y2)e�x

2�y

2

= 0

Observe that 8y, (0, y) is a solution to the system above.

f

xx

= 2y(2x4 � 5x2 + 1)e�x

2�y

2

;

f

yy

= 2x2y(2y2 � 3)e�x

2�y

2

;

f

xy

= 2x(x2 � 1)(2y2 � 1)e�x

2�y

2

Note that when x = 0, f

yy

(0, y) = 0and f

xy

(0, y) = 0.Graph of f(x, y) = x

2ye

�x

2�y

2

Thus, D = f

xx

(0, y)fyy

(0, y)� f

xy

(0, y) = 0.

Math 55 Relative and Absolute Extrema 9/ 15

Relative Extrema

Recall. If a continuous function f(x) has only one criticalnumber, then a relative maximum(minimum) has to be theabsolute maximum(minimum). However, this is not true for

f(x, y).

Consider f(x, y) = 3xey � x

3 � e

3y. The only critical point of fis (1, 0).

Observe from the graph that

f has a relative maximum at (1, 0) but it is not the absolutemaximum.

Math 55 Relative and Absolute Extrema 10/ 15

Absolute Extrema

Definition

Let R ⇢ R2. Then R is said to be

closed if it contains all its boundary points; open if itdoes not contain any of its boundary points

bounded if it can be contained in a disk (of some radius).

Theorem

If f(x, y) is continuous on some closed and bounded region

R ⇢ R2

, then there exist points (x1

, y

1

), (x2

, y

2

) 2 R such that

f(x1

, y

1

) � f(x, y) and f(x2

, y

2

) f(x, y) for all (x, y) 2 R, i.e.

f attains its absolute maximum and absolute minimum in R at

(x1

, y

1

) and (x2

, y

2

), resp.

Remark. If f has an absolute extremum at (x1

, y

1

), then(x

1

, y

1

) is either a critical point or a boundary point of f .Math 55 Relative and Absolute Extrema 11/ 15

Absolute Extrema

Example

Find the absolute minimum and maximum values off(x, y) = x

2 � 2xy + 2y on the triangular region D with vertices(0, 0), (4, 0) and (0, 2).

Solution. We first find the critical points of f . The system(f

x

= 2x� 2y = 0

f

y

= �2x+ 2 = 0has a unique solution (1, 1) 2 D.

Next, consider the boundary of D composed of 3 line segments:

1 2 3 4

1

2

(1, 1)

l2

l1

l3

On l

1

, y = 0. Define g

1

(x) = f(x, 0) = x

2.Note that g(x) attains its extremum wheng

0(x) = 2x = 0, i.e. at the point (0, 0) andat the other endpoint (4, 0).

Math 55 Relative and Absolute Extrema 12/ 15

Absolute Extrema

On l

2

, x = 0. Define g

2

(y) = f(0, y) = 2y. Note that g2

(y)attains its extremum at the endpoints (0, 0) and (0, 2).

On l

3

, y = 4�x

2

. Define

g

3

(x) = x

2 � 2x

✓4� x

2

◆+ 2

✓4� x

2

= 2x2 � 5x+ 4

It attains its extremum wheng

0(x) = 4x� 5 = 0, i.e. at�5

4

,

11

8

�.

So the candidates for the absolute extrema are: (1, 1), (0, 0),(4, 0), (0, 2) and

�5

4

,

11

8

f(1, 1) = 1f(0, 0) = 0

f(4, 0) = 16f(0, 2) = 4

f

�5

4

,

11

8

�= 7

8

Hence f has a maximum value of 16 at (4, 0) and minimumvalue of 0 at (0, 0).

Math 55 Relative and Absolute Extrema 13/ 15

Exercises

1 Find and classify all critical points ofa. f(x, y) = x

3y + 12x2 � 8y

b. f(x, y) = e

x cos y

c. f(x, y) = xy + x

�1 + y

�1

d. f(x, y) = (x2 + y

2)ey2�x

2

2 Show that⇣±1,±

p2

2

⌘are critical points of

f(x, y) = x

2

ye

�x

2�y

2

and classify each of them.

3 Find the points on the surface y

2 = 9 + xz that are closestto the origin. What is the distance of the surface from theorigin?

4 Find the absolute extrema of f(x, y) = 2x3 + y

4 over theunit disk centered at the origin.

5 Find the dimension of the rectangular box with largestvolume if the total surface area is 64 cm2.

Math 55 Relative and Absolute Extrema 14/ 15

References

1 Stewart, J., Calculus, Early Transcendentals, 6 ed., ThomsonBrooks/Cole, 2008

2 Leithold, L., The Calculus 7

3 Dawkins, P., Calculus 3, online notes available athttp://tutorial.math.lamar.edu/

Math 55 Relative and Absolute Extrema 15/ 15

Recommended