Calculus

You are given a piece of sheet metal that is twice as long as it is wide an has an area of 800m^2. Find the dimensions of the rectangular box that would contain a max volume if it were constructed from this piece of metal by cutting out squares of equal area at all four corners and folding up all the sides. The box will not have a lid.

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  1. First things first:

    width --- s
    length ---- 2s

    2s^2 = 800
    s^2 = 400
    s = 20

    so the piece of metal is 20 by 40

    let the side of the square to be cut out be x
    so the width is 20-2x
    the length is 40-2x
    the height is x

    Volume = x(20-2x)(40-2x)
    = 2x^3 - 120x^2 + 800x
    d(Volume)/dx = 6x^2 - 240x + 800 = 0 for a max/min of Volume

    3x^2 - 120x + 400 = 0

    use the formula to solve for x
    remember that x must be between 0 and 10 or else you get a negative side.
    (hope you get appr 3.67

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  2. messed up in my expansion

    volume should have been
    4x^3- 120x^2 + 800x

    and V' = 12x^2 - 240x + 800 = 0
    3x^2 - 60x + 200=0

    to get x = 4.23

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