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Thread: Min average cost

  1. #1
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    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?
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  2. #2
    Junior Member ignite's Avatar
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    Re: 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.
    Last edited by ignite; Apr 24th 2012 at 01:03 PM.
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  3. #3
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    Re: Min average cost

    I'm not sure I understand how to do that. I know its a simple problem I'm just lost.
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  4. #4
    Junior Member ignite's Avatar
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    Re: 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
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