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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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Cost A manufacturer has an order for 1000 wooden benches that can be produced at two locations. Let x1 and x2 be the numbers of wooden benches produced at the two locations. The cost function is modeled by

C = 0.25 x 1 2 + 10 x 1 + 0.15 x 2 2 + 12 x 2 .

Use Lagrange multipliers to find the number of wooden benches that should be produced at each location to minimize the cost.

To determine

To calculate: The number of wooden benches that should be produced at each location to minimize the cost by application of Lagrange multipliers when an order for 1000 wooden benches has to be produced at two different locations where x1 and x2 is assumed to be the number of wooden benches produced at two different locations and the cost function is given by,

C=0.25x12+10x1+0.15x22+12x2

Explanation

Given Information:

An order for 1000 wooden benches has to be produced at two different locations where x1 and x2 is assumed to be the number of wooden benches produced at two different locations and the cost function is given by the function,

C=0.25x12+10x1+0.15x22+12x2

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:

Here the function is C=0.25x12+10x1+0.15x22+12x2 and the subject to constraints is,

x1+x2=1000

Then, let f(x1,x2)=0.25x12+10x1+0.15x22+12x2 and g(x1,x2)=x1+x21000.

Then define a new function F as,

F(x1,x2,λ)=f(x1,x2)λg(x1,x2)=0.25x12+10x1+0.15x22+12x2λ(x1+x21000)

Now solve the equation by finding the partial derivatives of F with respect to x1,x2 and λ as,

Fx1(x1,x2,λ)=0.50x1+10λFx2(x1,x2,λ)=0.30x2+12λFλ(x1,x2,λ)=(x1+x21000)

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

0

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