Marginal Average Cost Function

Limits, differentiation, related rates, integration, trig integrals, etc.

Marginal Average Cost Function

This problem feels pretty straight forward, but I always like to double check. Thus far I have:

For all parts, consider the cost function:
$C(x)=500+4.5x+.01x^2$

A. Give the extra cost of producing one more unit when the production is already at a level of x=50
I gave $5.51 because C(51)-C(50)=5.51 B. Give the marginal cost function. I gave $M(x)=(.01x^2)-(.01(x-1)^2)$ C. Give the average marginal cost function. This one I'm having problems with (more so if I did the first two wrong), but I think it has something to do with a summation formula. I was thinking something along the lines of $f(x)=\frac{\Sigma(((.01x^2)-(.01(x-1)^2))+((.01x^2)-(.01(x-1)^2))...}{x}$ Or something... I really don't know. I think I'm pretty good with the whole margins thing and averages were never hard, but I'm looking at this one with a bit of a blank. After reading through the basic summations I'm getting that I need some n and n+1, but I'm not sure how to lay it out. GreenLantern Posts: 23 Joined: Sat Mar 07, 2009 10:47 pm Sponsor GreenLantern wrote:A. Give the extra cost of producing one more unit when the production is already at a level of x=50 I gave$5.51 because C(51)-C(50)=5.51

Sounds good to me.

GreenLantern wrote:B. Give the marginal cost function.
I gave
$M(x)=(.01x^2)-(.01(x-1)^2)$

Isn't the "marginal cost function" just the derivative of the cost function?

GreenLantern wrote:C. Give the average marginal cost function.
This one I'm having problems with (more so if I did the first two wrong), but I think it has something to do with a summation formula.

I think (from Exercise 3 here) that the "average marginal" (or "marginal average"?) cost function is the derivative of (the marginal form of) the average cost function. Possibly....

stapel_eliz

Posts: 1803
Joined: Mon Dec 08, 2008 4:22 pm

Re: Marginal Average Cost Function

Okay, maybe I missed that looking over my notes. So C(x) becomes...
$C'(x)=4.5+.02x$
... after the derivative becoming the Marginal Cost function.

As far as the Average Marginal Cost goes, I don't know what the C(x) with the line over it is, but I think I still got it.

First I take the cost function itself and put it over x.
$C(x)=\frac{C(x)}{x}\ OR\ C(x)=\frac{500+4.5x+.01x^2}{x}$
It can be read both ways? Assuming this is what makes the C(x) with the line over it.
Then we simplify it until we can take the derivative, thus making the Average Cost formula into the Marginal Average Cost formula.
$C(x)=\frac{500}{x}+4.5+.01x$
Becomes...
$C(x)=\500x^-^1+4.5+.01x$
Stop! Derivative time!
$C'(x)=\500x^-^2+.01$
Simplify...
$C(x)=\frac{500}{x^2}+.01$
And I suppose that is right...

So I need to find the average cost formula first then find the marginal cost formula from the average one. Because Marginal + Average = Marginal Average right?

Thanks a bunch!
GreenLantern

Posts: 23
Joined: Sat Mar 07, 2009 10:47 pm

GreenLantern wrote:So C(x) becomes...
$C'(x)=4.5+.02x$
... after the derivative becoming the Marginal Cost function.

That is my understanding, yes.

GreenLantern wrote:I don't know what the C(x) with the line over it is...

I believe that it is common notation indicating "average". For instance, if you have a list of x-values, the average of the list might be given as $\overline{x}.$

GreenLantern wrote:First I take the cost function itself and put it over x.
$C(x)=\frac{C(x)}{x}\ OR\ C(x)=\frac{500+4.5x+.01x^2}{x}$

As long as the $C(x)$ on the left-hand sides of the "equals" signs is meant to have the bar over it, yes. (The function itself is not equal to the function divided by x; hence, the different notation.)

GreenLantern wrote:Derivative time!
$\overline{C'(x)}=500x^-^2+.01$

Shouldn't there be a "minus" sign in front, from the -1 exponent?

stapel_eliz

Posts: 1803
Joined: Mon Dec 08, 2008 4:22 pm

Re:

stapel_eliz wrote:
GreenLantern wrote:Derivative time!
$\overline{C'(x)}=500x^-^2+.01$

Shouldn't there be a "minus" sign in front, from the -1 exponent?

What -1 exponent? I only have the -2 exponent... What I'm getting is I'm missing something during the change to a derivative, right? So... here.
$C(x)=\500x^-^1+4.5+.01x$

$C'(x)=-500x^-^2+.01$
Oh. I see it now! Yes, I missed multiplying the 500 by -1. I simply did 500x1=500. And it's all because I pay attention Thanks Liz!
GreenLantern

Posts: 23
Joined: Sat Mar 07, 2009 10:47 pm