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Thread: Optimizations problem with derivatives

  1. #1
    s3a
    s3a is offline
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    Optimizations problem with derivatives

    Question: (Attached) 59 b and c (I did a).

    Answers:
    b: About 8.5 inches by 2 inches.
    c: 20/sqrt(3) inches by 20*sqrt(2/3) inches

    My work is also attached for #59 a (Page 12/14 of pdf file) which I did successfully. I cannot complete b to get to c and for b my problem is that, I cannot find a relationship between the two variables to differentiate a function with one variable. As for #59 b, I thought I was on to something and wrote some unnecessary (I think) things.

    Any help would be greatly appreciated!
    Thanks in advance!
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  2. #2
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    I'm not sure if I'm correct, but after several mistakes I derived an answer for the width of the plank slightly different from $\displaystyle 8.5\,\mbox{in}$. I used the equations

    $\displaystyle \begin{aligned}
    y&=\sqrt{100-x^2}-\sqrt{10}\\
    A&=2xy=2x(\sqrt{100-x^2}-\sqrt{10}).
    \end{aligned}$

    Differentiating, I obtained

    $\displaystyle \begin{aligned}
    A'&=2\sqrt{100-x^2}+2x\cdot\frac{1}{2}(100-x^2)^{-\frac{1}{2}}\cdot(-2x)-2\sqrt{10}\\
    &=2\sqrt{100-x^2}-2x^2(100-x^2)^{-\frac{1}{2}}-2\sqrt{10}=0\quad\mbox{at max}.
    \end{aligned}$

    Adding $\displaystyle 2\sqrt{10}$, multiplying both sides by $\displaystyle \sqrt{100-x^2}$, and squaring gives us a quadratic equation from which I derived my final answer.
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