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Math Help - Marginal Cost derivative

  1. #1
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    Marginal Cost derivative

    I have an issue with a problem.

    I have the following function...

    Cost: C^2 = x^2 + 100(x)^[1/2] + 50

    I'm asked to find and interpret the marginal cost dC/dx at x = 5.

    Using implicit differentiation, which isn't really too hard because each variable is already on separate sides of the equation, I came up with the answer of

    x + 25(x)^{-1/2} = C.

    However, when I plug 5 into the derivative, I get an answer of about 16.18. I should be getting 0.94. Can someone give me a tip on what I've done wrong here?

    Thank you.
    Last edited by skeeter; September 27th 2012 at 04:08 AM.
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Marginal Cost derivative

    You neglected to use the chain rule when you differentiated. The right side of your equation after differentiation should be C\cdot\frac{dC}{dx}. Now solve for \frac{dC}{dx} by dividing through by C.
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    Re: Marginal Cost derivative

    Quote Originally Posted by MarkFL2 View Post
    You neglected to use the chain rule when you differentiated. The right side of your equation after differentiation should be C\cdot\frac{dC}{dx}. Now solve for \frac{dC}{dx} by dividing through by C.
    I don't quite understand what you mean...The way I THOUGHT it was supposed to be worked through as is as follows...

    c^2 = x^2 + 100(x)^{1/2} + 50

    = \frac{d}{dx}C = \frac {d}{dx} x^2 + \frac {d}{dx} 100(x)^{1/2} + \frac {d}{dx} 100

    = 2C \frac {dC}{dX} = 2x + \frac {50}{x^{1/2}}

    = \frac {dc}{dx} (C) = x + \frac {25}{x^{1/2}}

    When you say divide through by C what exactly do you mean by that? I thought that once the variables with \frac {dc}{dx} are isolated on one side, that you then factor that out, and then divide the x-valued side, by the C valued side. If I were to do this, wouldn't I end up with this?

    \frac {dC}{dx} = \frac {X}{C} + \frac {25C}{x^{1/2}}

    I didn't think this would be correct because I'm supposed to solve for C. Or now do I plug in x=5, THEN isolate C?

    Still a bit lost...
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  4. #4
    MHF Contributor MarkFL's Avatar
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    Re: Marginal Cost derivative

    You are given:

    C^2=x^2+100x^{\frac{1}{2}}+50

    Implicitly differentiating with respect to x, we find:

    2C\cdot\frac{dC}{dx}=2x+50x^{-\frac{1}{2}}

    \frac{dC}{dx}=\frac{x+25x^{-\frac{1}{2}}}{C}

    Now, when x=5, we have C=5\sqrt{4\sqrt{5}+3} hence:

    \frac{dC}{dx}\left|_{x=5}=\frac{5+5 \sqrt{5}}{5\sqrt{4\sqrt{5}+3}}=\frac{1+\sqrt{5}}{ \sqrt{4\sqrt{5}+3}}\approx0.936349095209
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    Re: Marginal Cost derivative

    Quote Originally Posted by MarkFL2 View Post

    Now, when x=5, we have C=5\sqrt{4\sqrt{5}+3} hence:[/TEX]
    Your post makes sense, but I can't see where this value of C is coming from?

    Also, wouldn't 25x^{-1/2} = 25(5)^{-1/2},

    and not 25x^{-1/2} = 5(5)^{-1/2}
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  6. #6
    MHF Contributor MarkFL's Avatar
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    Re: Marginal Cost derivative

    To find C, use:

    C=\sqrt{x^2+100x^{\frac{1}{2}}+50}

    25(5)^{-\frac{1}{2}}=\frac{25}{\sqrt{5}}=5\sqrt{5}
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  7. #7
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    Re: Marginal Cost derivative

    Quote Originally Posted by MarkFL2 View Post
    To find C, use:

    C=\sqrt{x^2+100x^{\frac{1}{2}}+50}

    25(5)^{-\frac{1}{2}}=\frac{25}{\sqrt{5}}=5\sqrt{5}
    Thanks so much for all the help (and patience)!
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  8. #8
    MHF Contributor MarkFL's Avatar
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    Re: Marginal Cost derivative

    It takes no patience at all to help someone who shows their work/thoughts and wants to understand the problem!
    Thanks from astuart
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