   Chapter 10.3, Problem 15E ### 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

# If the total cost function for a lamp is C ( x )   =   250   +   33 x +   0.1 x 2 dollars, producing how many units, x, will result in a minimum average cost per unit? Find the minimum average cost.

To determine

To calculate: The minimum average cost and the number of units produced that is x for the cost function for lamp is C(x)=250+33x+0.1x2.

Explanation

Given Information:

The provided cost function is:

C(x)=250+33x+0.1x2

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.

And the average cost is defined as:

C¯(x)=C(x)x

Where C(x) is the cost function.

Calculation:

Consider the cost function for lamp C(x)=250+33x+0.1x2.

Average cost is given as:

C¯(x)=C(x)x=250+33x+0.1x2x=250x+33+0.1x

The absolute maxima and absolute minima will occur only at the critical points. To calculate the critical points of the average cost function, find the first derivative of the function:

C¯(x)=250x+33+0.1xddx(C¯)=ddx(250x+33+0.1x)

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

R¯(x)=ddx(250x1)+ddx(33)+ddx(0

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