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Thread: Max/Min Of Area/Volume

  1. #1
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    Max/Min Of Area/Volume

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    Last edited by AlphaRock; Apr 1st 2009 at 01:45 AM.
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  2. #2
    Member arpitagarwal82's Avatar
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    let length and width of bas be x (its square so both are equal)


    and height be y.

    so volume $\displaystyle x^2y = 2$
    so$\displaystyle y = 10/x^2$

    area of sides and bottom = $\displaystyle x^2 + 4xy$
    area of top =$\displaystyle x^2$

    so cost
    $\displaystyle C = 10(x^2 + 4xy) + 20x^2$
    $\displaystyle = 30x^2 + 40xy$
    $\displaystyle = 30x^2 + 400/x$ (since $\displaystyle y = 10/x^2)$

    for minimum cost
    dC/dt = 0

    solve for x now.
    then solve for y ($\displaystyle y= 10/x^2$)
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