   Chapter 10, Problem 31RE ### 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

# In Problems 31-36, cost, revenue, and profit are in dollars and x is the number of units. C ( x ) = 3 x 2 + 15 x + 75 Cost Suppose the total cost function for a product isHow many units will minimize the average cost? Find the minimum average cost.

To determine

To calculate: The number of units to minimize the average cost where the cost function is C(x)=3x2+15x+75.

Explanation

Given Information:

The provided cost function is C(x)=3x2+15x+75.

Formula used:

To find relative maxima and minima of a function,

1. Set the first derivative of the function to zero, f(x)=0, to find the critical values of the function.

2. Substitute the critical values into f(x) and calculate the critical points.

3. Calculate the sign of the function to the left and right of the critical values.

(a) If f(x)<0 and f(x)>0 on the left and right side respectively of a critical value, then the point is relative minimum.

(b) If f(x)>0 and f(x)<0 on the left and right side respectively of a critical value, then the point is relative maximum.

Calculation:

Consider the cost function is C(x)=3x2+15x+75.

Since, the average cost function is C¯(x)=C(x)x.

C¯(x)=3x2+15x+75x=3x2x+15xx+75x=3x+15+75x

Thus, the average cost function is C¯(x)=3x+15+75x.

Calculate the first derivative of the function C¯(x)=3x+15+75x.

C¯(x)=375x2

Set the first derivative equal to 0.

375x2=0

Solve for x.

3=75x2x2=753x2=25

Take square root on both sides.

x2=25x=±5

Thus, the critical points are x=5 and x=5

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