   Chapter 14, Problem 56RE

Chapter
Section
Textbook Problem

Find the absolute maximum and minimum values of f on the set D.55. f ( x , y ) = e − x 2 − y 2 ( x 2 + 2 y 2 ) ; D is the disk x 2 + y 2 ≤ 4

To determine

To find: The absolute maximum and minimum values of the function f(x,y)=ex2y2(x2+2y2) on the disk x2+y24 .

Explanation

Given:

The function is f(x,y)=ex2y2(x2+2y2) on the disk x2+y24 .

Calculation:

Find the partial derivatives of the function f(x,y)=ex2y2(x2+2y2) as follows,

fx(x,y)=ex2y2(2x)+(x2+2y2)((2x)ex2y2)=2xex2y2(1x22y2)fy(x,y)=ex2y2(4y)+(x2+2y2)((2y)ex2y2)=2yex2y2(2x22y2)

Set the partial derivatives fx(x,y)=2xex2y2(1x22y2),fy(x,y)=2yex2y2(2x22y2) to zero and find the values of xandy .

2xex2y2(1x22y2)=0x=0orx2+2y2=1

2yex2y2(2x22y2)=0y=0

Possibility 1:

Assume, x=0 ,

fy(x,y)=2yex2y2(2x22y2)=0y=0orx2+2y2=2y=0ory=±1(x=0)

Thus, the critical points of the function f(x,y)=ex2y2(x2+2y2) are (0,0)and(0,±1) .

Substitute (0,0) in the function f(x,y)=ex2y2(x2+2y2) and obtain,

f(x,y)=ex2y2(x2+2y2)f(0,0)=e0(0+0)=1(0)=0

Thus, the value of f(0,0)=0 .

Substitute (0,±1) in the function f(x,y)=ex2y2(x2+2y2) and obtain,

f(x,y)=ex2y2(x2+2y2)f(0,±1)=e1(0+2(±1)2)=e1(2)=2e1

Thus, the value of f(0,±1)=2e1 .

This critical points (0,0)and(0,±1) are lie inside the given disk x2+y24 .

Substitute (0,0) in the inequality x2+y24 and check.

x2+y24(0)2+02404

Thus, the point satisfy the inequality x2+y24

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