# Math Help - Min average cost

1. ## Min average cost

The cost function for a company is C(x)=2000+96x+4x3/2 where x represents thousands of units. Is there a production level that minimizes average cost?

2. ## 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.

3. ## Re: Min average cost

I'm not sure I understand how to do that. I know its a simple problem I'm just lost.

4. ## 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