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Math Help - Max/Min Of Area/Volume

  1. #1
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    Max/Min Of Area/Volume

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    Last edited by AlphaRock; April 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 x^2y = 2
    so  y = 10/x^2

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

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

    for minimum cost
    dC/dt = 0

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