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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 73-78, use Lagrange multipliers to find the indicated extremum. Assume that the variables are positive.

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

Constraint: 2 x   +   y   +   z =   5

To determine

To calculate: The maximum value of the function f(x,y,z)=x2z+yz subject to constraint 2x+y+z=5 by the application of Lagrange multipliers.

Explanation

Given Information:

The provided function is f(x,y,z)=x2z+yz and the subject to constraints are 2x+y+z=5.

Formula used:

To find the maximum or minimum values of function f follow the steps as,

Step 1: Solve the provided system of equations as;

Fx(x,y,λ)=0Fy(x,y,λ)=0Fλ(x,y,λ)=0

Step 2: Next calculate the function f at each of the solution point obtained from first step where the greatest value obtained gives the maximum of function f subject to constraint g(x,y)=0 and the least value gives the minimum of function f subject to constraint g(x,y)=0.

Calculation:

First let f(x,y,z)=x2z+yz and g(x,y,z)=2x+y+z5.

Then define a new function F as;

F(x,y,z,λ)=f(x,y,z)λg(x,y,z)=x2z+yzλ(2x+y+z5)

Now solve the equation by finding the partial derivatives of F with respect to x, y, z and λ as;

Fx(x,y,z,λ)=2xz2λFy(x,y,z,λ)=zλFz(x,y,z,λ)=x2+yλFλ(x,y,z,λ)=(2x+y+z5)

Then set each of the obtained partial derivatives equal to 0 as;

2xz2λ=0zλ=0x2+yλ=0&

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