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calculus

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Suppose you have 400 meters of fencing and you wish to enclose a rectangular space with two parallel partitions: What is the maximum total area of the enclosed space, in square meters?

  • calculus - ,

    If the original rectangle is x m wide and y m long
    then the partitions are x m as well
    so we have
    2y + 4x = 400
    y + 2x = 200
    y = 200 - 2x

    Area = xy
    A = x(200-2x) = 200x - 2x^2
    dA/dx = 200 - 4x = 0 for a max of A
    200 = 4x
    x = 50
    then y = 200 - 100 = 100

    if x = 50 and y = 100, we get a maximum area
    which is xy or 50(100) = 5000 m^2

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