Chapter 7.6, Problem 3CP

### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

Chapter
Section

### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
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# The manufacturer in Example 2 now has $80,000 available for labor and capital. What is the maximum number of units that can be produced? To determine To calculate: The maximum production level when labour and capital cannot exceed$80000 if manufacture production function is f(x,y)=100x34y14.

Explanation

Given Information:

The Cob-Douglas function of the manufacture production is,

f(x,y)=100x34y14

x represents thee units of labour by represent the unit of capital also Given that the labour and the capital does not exceed $80000. Formula used: Step 1: Take the constraint using the given labour unit cost and capital unit cost and capital unit cost. Step 2: If f(x,y) be the function which have to maximize and. g(x,y) be the constraint then find new function. F(x,y,λ)=f(x,y)λg(x,y) Where λ is the langrage multiplier Step3: Finding the partial derivatives of f(x,yλ) with respect to x, y and λ. Step 4: Solve the equation and find the value of x and y equating the partial derivative with zero. Step 5: Putting that values of x and y in the given equation find the maximum production level Calculation: Consider the given equation, f(x,y)=100x34y14 Since each labour costs 200 and each capital costs$250 then the total labour and capital is 200x+250y.

As the labour and capital does not exceed \$80000 then,

250x+250y=80000150x+250y80000=0

Consider, g(x,y)=150x+250y80000

Then f(x,y)=f(x,y)λg(x,y)

100x34y14λ(150x+250y80000)

Now, The partial derivatives of F(x,y,λ) with respect to x, y, λ one,

Fy(x,y,λ)=100×34x341y14150λ=25x14y14150λ

And,

Fy(x,y,λ)=100×14yλ1x34250λ=25x34y34250λ

And Fλ(x,y,λ)=(150x+250y80000)

Now, equate all above equation to zero

Fx(x,y,λ)=0Fy(x,y,λ)=0Fλ(x,y,λ)=0

Then substituting the values,

75x14y14150λ=0

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