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Thread: Differentiation to get maximum point

  1. #1
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    Differentiation to get maximum point

    Hi! I've been trying to calculate the maximum point of a curve, y = xe^{0.39(1-x/1619)}-x, so I differentiated it and got dy/dx = e^{0.39(1-x/1619)}*(1-0.39/1619x) - 1, and I tried to equate it to zero and find the x, but I just don't know how to get it, can anyone please help?
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Differentiation to get maximum point

    We are given:

    $\displaystyle y=xe^{\Large{0.39\left(1-\frac{x}{1619}\right)}}-x$

    I would choose to write this as:

    $\displaystyle y=xe^{\Large{\frac{39}{161900}(1619-x)}}-x$

    And so equating the derivative to zero, we obtain:

    $\displaystyle \frac{dy}{dx}=e^{\Large{\frac{39}{161900}(1619-x)}}-\frac{39x}{161900}e^{\Large{\frac{39}{161900}(1619-x)}}-1=0$

    Or:

    $\displaystyle e^{\Large{\frac{39}{161900}(1619-x)}}(161900-39x)-161900=0$

    This cannot be solved algebraically using elementary functions, so using a numeric root finding technique, we find:

    $\displaystyle x\approx768.7939617019124301$

    We see that at $\displaystyle x=\frac{161900}{39}$ the derivative is negative, and at $\displaystyle x=0$ the derivative is positive, and so by the first-derivative test, we may conclude that the extremum is a maximum.
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  3. #3
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    Re: Differentiation to get maximum point

    Quote Originally Posted by MarkFL View Post
    We are given:

    $\displaystyle y=xe^{\Large{0.39\left(1-\frac{x}{1619}\right)}}-x$

    I would choose to write this as:

    $\displaystyle y=xe^{\Large{\frac{39}{161900}(1619-x)}}-x$

    And so equating the derivative to zero, we obtain:

    $\displaystyle \frac{dy}{dx}=e^{\Large{\frac{39}{161900}(1619-x)}}-\frac{39x}{161900}e^{\Large{\frac{39}{161900}(1619-x)}}-1=0$

    Or:

    $\displaystyle e^{\Large{\frac{39}{161900}(1619-x)}}(161900-39x)-161900=0$

    This cannot be solved algebraically using elementary functions, so using a numeric root finding technique, we find:

    $\displaystyle x\approx768.7939617019124301$

    We see that at $\displaystyle x=\frac{161900}{39}$ the derivative is negative, and at $\displaystyle x=0$ the derivative is positive, and so by the first-derivative test, we may conclude that the extremum is a maximum.
    YES!! THANK YOU SO MUCH! I used the Newton's method with r programming, took a while cuz I've forgotten everything :P
    Thanks from MarkFL
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