word problem

[kris17]

If C( x) is the total cost of producing x weather radios, then C'( x) represents:
A) The rate of change of units produced with respect to cost.
B) The average cost of producing x radios.
C) The approximate cost of producing one additional radio after the xth item is produced.
D) The number of radios to produce to minimize cost

I said A because the derivative tells us the rate of change of a function at a particular time, so A is the only one that says that.
Am I right?

 

[Deleted member 4993]

If C( x) is the total cost of producing x weather radios, then C'( x) represents:
A) The rate of change of units produced with respect to cost.
B) The average cost of producing x radios.
C) The approximate cost of producing one additional radio after the xth item is produced.
D) The number of radios to produce to minimize cost

I said A because the derivative tells us the rate of change of a function at a particular time, so A is the only one that says that.
Am I right?

However, you do not have a function of time. You have a function of x - number of units produced.

The derivative gives you the rate of change of the dependent variable with respect to the independent variable.

In this problem what is the independent variable - the number of radios produced.

In this problem what is the dependent variable - the cost of producing radios.

With this information, rethink your answer...

 

[soroban]

Hello, kris17!

This is a tricky one . . .

If \(\displaystyle C(x)\) is the total cost of producing \(\displaystyle x\) weather radios, then \(\displaystyle C'(x)\) represents:

(A) The rate of change of units produced with respect to cost. . No

Read it again.

"The rate of change of \(\displaystyle x\) with respect to \(\displaystyle C.\)"

This refers to: .\(\displaystyle \dfrac{dx}{dC}\) . . . and not \(\displaystyle C'(x) \,=\,\dfrac{dC}{dx}\)

(B) The average cost of producing \(\displaystyle x\) radios. . No

This would be: .\(\displaystyle \dfrac{C(x)}{x}\)

(C) The approximate cost of producing one additional radio after the \(\displaystyle x^{th}\) item is produced.

I had to read this a few times to get it.

\(\displaystyle C'(x) \,=\,\dfrac{dC}{dx}\)

This is "The rate of change of Cost with respect to number of Radios."


Suppose \(\displaystyle x = 7.\)

Then \(\displaystyle C'(7)\) is the rate of change of Cost when 7 radios are produced.

It approximates the increase in Cost to produce the 8th radio.

Get it?

(D) The number of radios to produce to minimize cost. . No

This would be a value of \(\displaystyle x.\)

We would find this value by solving: .\(\displaystyle C'(x) \,=\,0\)

 

[JeffM]

If C( x) is the total cost of producing x weather radios, then C'( x) represents:
A) The rate of change of units produced with respect to cost.
B) The average cost of producing x radios.
C) The approximate cost of producing one additional radio after the xth item is produced.
D) The number of radios to produce to minimize cost

I said A because the derivative tells us the rate of change of a function at a particular time, so A is the only one that says that.
Am I right?

C(x) is a total COST function. The DEPENDENT variable is total cost; the independent variable is units produced.

\(\displaystyle So\ \dfrac{dC}{dx} = C'(x) = the\ rate\ of\ change\ in\ total\ cost\ with\ respect\ to\ units\ produced.\)

But calculus deals with infinitessimal changes, and it makes no practical sense to think of the cost of 1/10¹⁰⁰ of a radio.

So at any non-negative, integer value of x, the derivative approximates the incremental cost of producing one more radio or the incremental reduction in cost of producing one fewer radio. The correct answer is (C), but the question is horribly worded. The question should make clear that it is talking about incremental (or, in the jargon of economics, marginal) cost.

By the way, rate of change does not always mean change as a function of time. Here it means change as a function of units produced.

PS I did not explain the math involved

\(\displaystyle \Delta C \approx \dfrac{dC}{dx} * \Delta x = C'(x) * \Delta x \implies \Delta C \approx C'(x)\ if\ \Delta x = 1.\)

\(\displaystyle \Delta x = 1 \implies producing\ one\ more\ unit.\)

In terms of economic jargon:

\(\displaystyle C(x) = total\ cost\ function \implies \dfrac{C(x)}{x} = average\ cost\ function\ and\ C'(x) = marginal\ cost\ function.\)