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Math Help - finding local max of function via 1st derivative

  1. #1
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    finding local max of function via 1st derivative

    y=x(1-x)^(1/3)

    The derivative of the function turns out to be (1-x)^(1/3) - x/(3(1-x)^2/3)
    Using the calculator I found out the max is 0.750, .472...But when I set the derivative equal to 0...and tried to factor I got x=-3/11, x=1...I understand how to find the mix and max, my question is how to factor this particular equation...
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  2. #2
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    When you set the derivative equal to zero, you need to do the following algebra to get the critical point:

    0=(1-x)^{\frac{1}{3}}-\frac{x}{3(1-x)^{\frac{2}{3}}}

    (1-x)^{\frac{1}{3}}=\frac{x}{3(1-x)^{\frac{2}{3}}}

    1-x=\frac{x^3}{27(1-x)^2}

    27(1-x)^3=x^3

    3(1-x)=x

    3=4x

    x=.75
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  3. #3
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by bcahmel View Post
    y=x(1-x)^(1/3)

    The derivative of the function turns out to be (1-x)^(1/3) - x/(3(1-x)^2/3)
    Using the calculator I found out the max is 0.750, .472...But when I set the derivative equal to 0...and tried to factor I got x=-3/11, x=1...I understand how to find the mix and max, my question is how to factor this particular equation...
    Is...

    \displaystyle y^{'}= (1-x)^{\frac{1}{3}} - \frac{x}{3}\ (1-x)^{-\frac{2}{3}} = (1-x)^{\frac{1}{3}}\ \{1 - \frac{x}{3\ (1-x)}}\} (1)

    ... so that the maximim is for x=\frac{3}{4} ...



    Merry Christmas from Italy


    \chi \sigma
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  4. #4
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    Hello, bcahmel!

    y\:=\:x(1-x)^{\frac{1}{3}}

    \text{The derivative turns out to be: }\:y' \:=\: (1-x)^{\frac{1}{3}} - \dfrac{x}{3(1-x)^{\frac{2}{3}}}

    \text{Using the calculator I found out the max is: }\:(0.75,\;0.47247\hdots)

    We have: . (1-x)^{\frac{1}{3}} - \dfrac{x}{3(1-x)^{\frac{2}{3}}} \;=\;0


    Assuming x \ne 1, multiply through by 3(1-x)^{\frac{2}{3}}\!:

    . . 3(1-x) - x \;=\;0 \quad\Rightarrow\quad 3 - 3x - x \:=\:0

    . . -4x \:=\:-3 \quad\Rightarrow\quad x \:=\:\dfrac{3}{4}

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  5. #5
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    thanks to everyone for the help! I appreciate it greatly
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