# Choose the one alternative that best completes the statement or answers the question. A...

## Question:

Choose the one alternative that best completes the statement or answers the question.

A manufacturer has determined that the total cost c of producing q units of a product is given by {eq}c= 0.04q^2+4q+6400. {/eq} The average cost will be a minimum at a production level of _____

A) 100 units

B) 200 units

C) 400 units

D) 800 units

E) none of the above

## Optimizing Functions:

Optimizing functions entail that we can find the values at which the function is at a maximum or minimum. We can do this with the help of differentiation to search for the critical points of the function. The critical point of a function can be a location where the maxima or minima is observed.

## Answer and Explanation:

Determine the production level, *q*, that minimizes the average cost of the manufacturer. We determine the answer by first finding the average cost function, {eq}\displaystyle c_{ave} {/eq}, then taking its derivative, {eq}\displaystyle c'_{ave} {/eq}, equating it to zero, and solving for its root. Proceed with the solution in a step-by-step manner.

{eq}\begin{align} \displaystyle c &= 0.04q^2+4q+6400\\ \text{Take the average cost function}\\ \text{by dividing the cost function by q.}\\ c_{ave} &=\frac{0.04q^2+4q+6400}{q}\\ c_{ave} &= 0.04q+4+ \frac{6400}{q}\\ \text{Take the derivative.}\\ c'_{ave} &= \frac{d}{dq} \left(0.04q+4+ \frac{6400}{q}\right)\\ c'_{ave} &= 0.04-\frac{6400}{q^2}\\ \text{Equate it to zero.}\\ 0 &= 0.04-\frac{6400}{q^2}\\ 0.04 &= \frac{6400}{q^2}\\ q^2 &= \frac{6400}{0.04}\\ \text{Take the positive root.}\\ q &= \sqrt{ \frac{6400}{0.04}}\\ q&=\boxed{\rm 400\ units} \end{align} {/eq}

Therefore, the answer is {eq}\displaystyle \boxed{C)} {/eq}.

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#### Learn more about this topic:

from Math 104: Calculus

Chapter 9 / Lesson 3