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Math Help - find derivative, set to zero (partially solved...maybe)

  1. #1
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    find derivative, set to zero (partially solved...maybe)

    I need to find the maximum of this equation by finding the derivative and setting it to zero:

    f(x) = (1+20\sqrt{t})e^{-.05t} = e^{-.05t} + 20\sqrt{t}e^{-.05t}

    Now, solving for the derivative using the chain rule for the first operand, and using product rule+chain rule for the second operand, I get the following:

    f'(x) = \frac{10-t}{\sqrt{t}e^{.05t}} - \frac{.05}{e^{.05t}} = 0

    get common denominators:

     = \frac{10-t}{\sqrt{t}e^{.05t}} - \frac{.05\sqrt{t}}{\sqrt{t}e^{.05t}}

    Assuming that is correct so far, where do I go from there? Do I multiply through to cancel the denominator to get the following:

     10 - t - .05\sqrt{t} = 0

    If that's correct, then what do I do?
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  2. #2
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    Quote Originally Posted by sgcb View Post
    I need to find the maximum of this equation by finding the derivative and setting it to zero:

    f(x) = (1+20\sqrt{t})e^{-.05t} = e^{-.05t} + 20\sqrt{t}e^{-.05t}

    Now, solving for the derivative using the chain rule for the first operand, and using product rule+chain rule for the second operand, I get the following:

    f'(x) = \frac{10-t}{\sqrt{t}e^{.05t}} - \frac{.05}{e^{.05t}} = 0

    get common denominators:

     = \frac{10-t}{\sqrt{t}e^{.05t}} - \frac{.05\sqrt{t}}{\sqrt{t}e^{.05t}}

    Assuming that is correct so far, where do I go from there? Do I multiply through to cancel the denominator to get the following:

     10 - t - .05\sqrt{t} = 0

    If that's correct, then what do I do?
    let u = \sqrt{t} ...

    10 - u^2 - .05u = 0

    u^2 + .05u - 10 = 0

    u = \frac{-.05 \pm \sqrt{40.0025}}{2}

    since \sqrt{t} > 0 ...

    \sqrt{t} = \frac{-.05 + \sqrt{40.0025}}{2}

    t \approx 9.843

    max f(t) \approx 38.969
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  3. #3
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    Oh, cool! It never occurred to me to set it up that way, thanks skeeter! That was also the answer I got on the calculator.
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