   Chapter 10.3, Problem 30E ### Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042

#### Solutions

Chapter
Section ### Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042
Textbook Problem

# A firm can produce 100 units per week. If its total cost function is C   =   500   +   1500 x dollars and its total revenue function is R =   1600 x   — x 2 dollars, how many units, x, should it produce to maximize its profit? Find the maximum profit.

To determine

To calculate: The maximum profit of the provided cost function C(x)=500+1500x and the unit for which it is maximum.

Explanation

Given Information:

The provided cost function is:

C(x)=500+1500x

And the revenue function is:

R(x)=1600xx2

Formula used:

If f(x) and g(x) are two differentiable functions then by the property of derivative:

ddx(f(x)+g(x))=ddxf(x)+ddxg(x)

And,

ddxxn=nxn1

Where, n is a constant and x is the variable.

If C(x) and R(x) are the cost and the revenue functions then the profit P(x) is defined as:

P(x)=R(x)C(x)

To evaluate the relative maxima and minima of a function,

Step 1: Evaluate the critical values of the function.

Step 2: Now substitute the critical values into f(x) to find the critical points.

Step 3: Evaluate f(x) at each critical value for which f(x)=0.

If f(a)<0, a relative maximum occurs at a.

If f(a)>0, a relative minimum occurs at a.

If f(a)=0, or f(a) is undefined, the second derivative test fails, now use the first derivative test.

Calculation:

Consider the provided cost function:

C(x)=500+1500x

And revenue function:

R(x)=1600xx2

Profit function can be written as:

P(x)=R(x)C(x)

Put provided values of cost and the revenue functions:

P(x)=1600xx2(500+1500x)=500+100xx2

The absolute maxima and absolute minima will occur only at the critical points

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