   Chapter 14.8, Problem 21E

Chapter
Section
Textbook Problem

Find the extreme values of f on the region described by the inequality.21. f(x, y) = x2 + y2 + 4x - 4y, x2 + y2 ≤ 9

To determine

To find: The extreme values of the function f(x,y)=x2+y2+4x4y on the disk x2+y29 .

Explanation

Given:

The function is f(x,y,z)=x2+y2+4x4y on the disk x2+y29 .

Definition used:

“The Lagrange multipliers defined as f(x,y,z)=λg(x,y,z) . This equation can be expressed as fx=λgx , fy=λgy , fz=λgz and g(x,y,z)=k ”.

Calculation:

Find the partial derivatives of the function f(x,y)=x2+y2+4x4y as follows,

f(x,y)=x2+y2+4x4yfx(x,y)=2x+4fy(x,y)=2y4

Set the partial derivatives fx(x,y)=2x+4,fy(x,y)=2y4 to zero and find the values of xandy .

fx(x,y)=2x+4=0x=2fy(x,y)=2y4=0y=2

Thus, the critical point of the function f(x,y)=x2+y2+4x4y is (2,2) .

Substitute (2,2) in the function f(x,y)=x2+y2+4x4y and obtain,

f(x,y)=x2+y2+4x4yf(2,2)=(2)2+(2)2+4(2)4(2)=4+488=8

Thus, the value of f(2,2)=8 .

This critical point (2,2) is lie inside the given disk x2+y29 .

Substitute (2,2) in the inequality x2+y29 and check.

x2+y29(2)2+2294+4989

Thus, the point satisfy the inequality x2+y29 .

Let the boubndary function of the inequality x2+y29 be g(x,y)=x2+y2=9 .

The Lagrange multipliers f(x,y)=λg(x,y) is computed as follows,

f(x,y)=λg(x,y)fx,fy=λgx,gyfx(x2+y2+4x4y),fy(x2+y2+4x4y)=λgx(x2+y2),gy(x2+y2)2x+4,2y4=λ2x,2y

Thus, the value of f(x,y)=λg(x,y) is 2x+4,2y4=λ2x,2y

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