   Chapter 7.6, Problem 14E ### 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 13-18, use Lagrange multipliers to find the indicated extremum. Assume that x, y, and z are positive. See Example 1. Maximize  f ( x , y , z ) = x 2 y 2 z 2 Constraint : x 2 + y 2 + z 2 − 9 = 0

To determine

To calculate: The indicated extremum of the function f(x,y,z)=x2y2z2 subject to constraint x2+y2+z29=0 by the use of language multipliers method.

Explanation

Given Information:

The objective function f(x,y,z)=x2y2z2 to maximize and the constraint x2+y2+z29=0.

Formula used:

To solve a function by lagrangean multiplier,

Step 1: Write the constraints in the form g(x,y)=0 and get g(x,y)

Step 2: Make new function

F(x,y,λ)=f(x,y)λg(x,y)

Step 3: Find the values of,

Fx(x,y,λ), Fy(x,y,λ), Fλ(x,y,λ)

Step 4: Solve the following system of equations

Fx(x,y,λ)=0Fy(x,y,λ)=0Fλ(x,y,λ)=0

Step 5: Evaluate f at each solution point obtained in the 4th step

The greatest value yields the maximum of f subject to the constraint g(x,y)=0 and the last

value yields the minimum of f subject to the constraint g(x,y)=0

Calculation:

Consider the function, f(x,y,z)=x2y2z2 and g(x,y)=x2+y2+z29=0.

The provided function and the constraints function can be write as new function by introducing a λ which is a langrange multiplier.

F(x,y,z,λ)=x2y2z2λ(x2+y2+z29)

Differentiate the above function with respect to x.

x[F(x,y,z,λ)]=x[x2y2z2λ(x2+y2+z29)]Fx(x,y,z,λ)=2xy2z2λ(2x)

Equate the above partial derivative function to 0 to get the value of λ.

Fx(x,y,z,λ)=02xy2z2λ(2x)=0λ(2x)=2xy2z2λ=y2z2

Differentiate the above function with respect to y.

y[F(x,y,z,λ)]=y[x2y2z2λ(x2+y2+z29)]Fy(x,y,z,λ)=2x2yz2λ(2y)

Equate the above partial derivative function to 0.

Fy(x,y,z,λ)=02x2yz2λ(2y)=02x2yz2=λ(2y)

Substitute y2z2 for λ

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