# Marginal Cost derivative

• September 26th 2012, 10:14 PM
astuart
Marginal Cost derivative
I have an issue with a problem.

I have the following function...

$Cost: C^2 = x^2 + 100(x)^[1/2] + 50$

I'm asked to find and interpret the marginal cost dC/dx at x = 5.

Using implicit differentiation, which isn't really too hard because each variable is already on separate sides of the equation, I came up with the answer of

$x + 25(x)^{-1/2} = C$.

However, when I plug 5 into the derivative, I get an answer of about 16.18. I should be getting 0.94. Can someone give me a tip on what I've done wrong here?

Thank you.
• September 26th 2012, 10:22 PM
MarkFL
Re: Marginal Cost derivative
You neglected to use the chain rule when you differentiated. The right side of your equation after differentiation should be $C\cdot\frac{dC}{dx}$. Now solve for $\frac{dC}{dx}$ by dividing through by $C$.
• September 26th 2012, 10:37 PM
astuart
Re: Marginal Cost derivative
Quote:

Originally Posted by MarkFL2
You neglected to use the chain rule when you differentiated. The right side of your equation after differentiation should be $C\cdot\frac{dC}{dx}$. Now solve for $\frac{dC}{dx}$ by dividing through by $C$.

I don't quite understand what you mean...The way I THOUGHT it was supposed to be worked through as is as follows...

$c^2 = x^2 + 100(x)^{1/2} + 50$

$= \frac{d}{dx}C = \frac {d}{dx} x^2 + \frac {d}{dx} 100(x)^{1/2} + \frac {d}{dx} 100$

$= 2C \frac {dC}{dX} = 2x + \frac {50}{x^{1/2}}$

$= \frac {dc}{dx} (C) = x + \frac {25}{x^{1/2}}$

When you say divide through by $C$ what exactly do you mean by that? I thought that once the variables with $\frac {dc}{dx}$ are isolated on one side, that you then factor that out, and then divide the x-valued side, by the C valued side. If I were to do this, wouldn't I end up with this?

$\frac {dC}{dx} = \frac {X}{C} + \frac {25C}{x^{1/2}}$

I didn't think this would be correct because I'm supposed to solve for C. Or now do I plug in x=5, THEN isolate C?

Still a bit lost...
• September 26th 2012, 11:13 PM
MarkFL
Re: Marginal Cost derivative
You are given:

$C^2=x^2+100x^{\frac{1}{2}}+50$

Implicitly differentiating with respect to $x$, we find:

$2C\cdot\frac{dC}{dx}=2x+50x^{-\frac{1}{2}}$

$\frac{dC}{dx}=\frac{x+25x^{-\frac{1}{2}}}{C}$

Now, when $x=5$, we have $C=5\sqrt{4\sqrt{5}+3}$ hence:

$\frac{dC}{dx}\left|_{x=5}=\frac{5+5 \sqrt{5}}{5\sqrt{4\sqrt{5}+3}}=\frac{1+\sqrt{5}}{ \sqrt{4\sqrt{5}+3}}\approx0.936349095209$
• September 26th 2012, 11:37 PM
astuart
Re: Marginal Cost derivative
Quote:

Originally Posted by MarkFL2

Now, when $x=5$, we have $C=5\sqrt{4\sqrt{5}+3}$ hence:[/TEX]

Your post makes sense, but I can't see where this value of C is coming from?

Also, wouldn't $25x^{-1/2} = 25(5)^{-1/2}$,

and not $25x^{-1/2} = 5(5)^{-1/2}$
• September 26th 2012, 11:48 PM
MarkFL
Re: Marginal Cost derivative
To find $C$, use:

$C=\sqrt{x^2+100x^{\frac{1}{2}}+50}$

$25(5)^{-\frac{1}{2}}=\frac{25}{\sqrt{5}}=5\sqrt{5}$
• September 26th 2012, 11:55 PM
astuart
Re: Marginal Cost derivative
Quote:

Originally Posted by MarkFL2
To find $C$, use:

$C=\sqrt{x^2+100x^{\frac{1}{2}}+50}$

$25(5)^{-\frac{1}{2}}=\frac{25}{\sqrt{5}}=5\sqrt{5}$

Thanks so much for all the help (and patience)!
• September 27th 2012, 12:00 AM
MarkFL
Re: Marginal Cost derivative
It takes no patience at all to help someone who shows their work/thoughts and wants to understand the problem! :)