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?

Printable View

- Apr 24th 2012, 12:52 PMmonicajoyce09Min 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? - Apr 24th 2012, 12:59 PMigniteRe: Min average cost
Average cost is defined by $\displaystyle AC(x)=\frac{C(x)}{x}$.

In order to minimise $\displaystyle AC(x)$ , find $\displaystyle x_0$ such that $\displaystyle AC'(x_0)=0$ and $\displaystyle AC''(x_0)>0$

You will find that $\displaystyle x_0=100$ satisfies these conditions. - Apr 24th 2012, 01:02 PMmonicajoyce09Re: Min average cost
I'm not sure I understand how to do that. I know its a simple problem I'm just lost.

- Apr 24th 2012, 01:12 PMigniteRe: Min average cost
$\displaystyle AC(x)=\frac{2000}{x}+96+4x^\frac{1}{2}$

Differentiating wrt x,

$\displaystyle AC'(x)=\frac{-2000}{x^2}+2x^\frac{-1}{2}$

$\displaystyle AC'(x)=0 \Rightarrow 2x^\frac{-1}{2}=\frac{2000}{x^2} \Rightarrow x_0=100$

Similarly find $\displaystyle AC''(x)$ and show that it's value at $\displaystyle x_0=100$ is greater than 0.

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