   Chapter 14, Problem 59RE

Chapter
Section
Textbook Problem

Use Lagrange multipliers to find the maximum and minimum values of f subject to the given constraint(s).59. f(x, y) = x2y; x2 + y2 = 1

To determine

To find: The extreme values of the function f(x,y)=x2y subject to the constraint x2+y2=1 by using Lagrange multipliers.

Explanation

Given:

The function f(x,y)=x2y which is subject to the constraint g(x,y)=x2+y2=1 .

Result 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:

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

f(x,y)=λg(x,y)fx,fy=λgx,gyfx(x2y),fy(x2y)=λgx(x2+y2),gy(x2+y2)2xy,x2=λ2x,2y

Thus, the value of f(x,y)=λg(x,y) is 2xy,x2=λ2x,2y .

The result 2xy,x2=λ2x,2y , can be expressed as follows,

2xy=λ(2x) (1)

x2=λ(2y) (2)

The extreme values of the function f(x,y)=3x+y is computed as follows,

For the equation (1), there are two possibilities such as either x=0 or y=λ .

Suppose x=0 , then the equation g(x,y)=x2+y2=1 ,

x2+y2=10+y2=1y=±1

Thus, the points are (0,1)and(0,1) .

Substitute (x,y)=(0,±1) in the function f(x,y)=x2y ,

f(x,y)=x2yf(0,±1)=(0)2(±1)=0

Thus, the value of f(0,±1)=0

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