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Math Help - finding minimum

  1. #1
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    finding minimum

     AC=\frac{TC}{x}=\frac{2000}{x}+20+\frac{10}{\sqrt{  x}}
    So i have to show that there is no minimum.

    thus i have to differentiate the function, which apparently gives this

     \frac{dAC}{dx}=AC'= -\frac{2000}{x^2}-\frac{5}{\sqrt[3]{}x^3}

    I don't know how they got that answer , can someone please explain?
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  2. #2
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    Quote Originally Posted by el123 View Post
     AC=\frac{TC}{x}=\frac{2000}{x}+20+\frac{10}{\sqrt{  x}}
    So i have to show that there is no minimum.

    thus i have to differentiate the function, which apparently gives this

     \frac{dAC}{dx}=AC'= -\frac{2000}{x^2}-\frac{5}{\sqrt[3]{}x^3}

    I don't know how they got that answer , can someone please explain?

    AC = \frac{2000}{x}+20+\frac{10}{\sqrt{x}}

    AC = 2000x^{-1} + 20 + 10x^{-\frac{1}{2}}

    use the power rule ...

    AC' = -2000x^{-2} - 5x^{-\frac{3}{2}}

    rewrite ...

    AC' = -\frac{2000}{x^2}-\frac{5}{\textcolor{red}{\sqrt{x^3}}}

    note there is no "cube" root in the denominator of the last term.
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