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Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

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Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
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Using Lagrange Multipliers In Exercises 1-12, use Lagrange multipliers to find the indicated extremum. Assume that x and y are positive. See Example 1.

Maximize f ( x , y ) = x 2 y 2

Constraint: 2 y x 2 = 0

To determine

To calculate: The maximum value of the function f(x,y)=x2y2 subject to the constraint 2yx2=0 by the use of Lagrange multipliers.

Explanation

Given Information:

The provided function is f(x,y)=x2y2 subject to the constraint 2yx2=0. Consider that the variables x and y are positive.

Formula used:

Method of Lagrange multipliers,

If the function f(x,y) contains a maximum or minimum subject to the constraint g(x,y)=0 then the maximum or minimum can occur at one of the critical numbers of the function F is,

F(x,y,λ)=f(x,y)λg(x,y) where, λ is a Lagrange multiplier.

Steps to determine the minimum or maximum of the function f.

1. Solve the system of equations,

Fx(x,y,λ)=0Fy(x,y,λ)=0Fλ(x,y,λ)=0

2. Determine the value of the function f at each solution obtained from the step 1.

The largest value gives the maximum value of function f subject to the constraint g(x,y)=0 and the lowest value gives the minimum value of function f subject to the constraint g(x,y)=0.

Calculation:

Consider the function,

f(x,y)=x2y2

The provided constraint is 2yx2=0.

So, g(x,y)=2yx2

Now, the new function F is,

F(x,y,λ)=x2y2λ(2yx2)

Find the partial derivative of function F to determine the critical number for F.

The partial derivatives of F with respect to x and then set it to zero.

Fx(x,y,λ)=x[x2y2λ(2yx2)]0=2x+2λx0=2x(1+λ)λ=1

The value x=0 does not gives maximum value for the function

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