   Chapter 7.6, Problem 10E ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

#### Solutions

Chapter
Section ### 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.Minimize f ( x , y ) = x 2 + y 2 Constraint: 2 x + 4 y − 15 = 0

To determine

To calculate: The minimum value of the function f(x,y)=x2+y2 subject to the constraint 2x+4y15=0 by the use of Lagrange multipliers.

Explanation

Given Information:

The provided function is f(x,y)=x2+y2 subject to the constraint 2x+4y15=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)=x2+y2

The provided constraint is 2x+4y15=0.

So, g(x,y)=2x+4y15

Now, the new function F is,

F(x,y,λ)=x2+y2λ(2x+4y15)

Differentiate above equation with respect to x, y and λ,

Fx(x,y,λ)=2x2x2+y22λFy(x,y,λ)=2y2x2+y24λ

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