Min average cost

Printable View

April 24th 2012, 12:52 PM
monicajoyce09

Min average cost

The cost function for a company is C(x)=2000+96x+4x^3/2 where x represents thousands of units. Is there a production level that minimizes average cost?
April 24th 2012, 12:59 PM
ignite

Re: Min average cost

Average cost is defined by $AC(x)=\frac{C(x)}{x}$ .
In order to minimise $AC(x)$ , find $x_0$ such that $AC'(x_0)=0$ and $AC''(x_0)>0$

You will find that $x_0=100$ satisfies these conditions.
April 24th 2012, 01:02 PM
monicajoyce09

Re: Min average cost

I'm not sure I understand how to do that. I know its a simple problem I'm just lost.
April 24th 2012, 01:12 PM
ignite

Re: Min average cost

$AC(x)=\frac{2000}{x}+96+4x^\frac{1}{2}$
Differentiating wrt x,
$AC'(x)=\frac{-2000}{x^2}+2x^\frac{-1}{2}$
$AC'(x)=0 \Rightarrow 2x^\frac{-1}{2}=\frac{2000}{x^2} \Rightarrow x_0=100$
Similarly find $AC''(x)$ and show that it's value at $x_0=100$ is greater than 0.

If you are new to Differentiation,please read Derivative - Wikipedia, the free encyclopedia

All times are GMT -8. The time now is 11:41 PM.

Copyright © 2005-2013 Math Help Forum. All rights reserved.

Copyright © 2005-2013 Math Help Forum. All rights reserved.