Math

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24 m of fencing are available to enclose a play area. what would the maximum area be if the fencing only went on 3 sides because the wall is used as the 4th side.
Please show you work

  • Math -

    l + 2w = perimeter
    l + 2w = 24
    l = 24-2w

    Area = l times w

    (24-2w)w = Area.
    24w -2w^2 = Area.

    If you are doing this using calculus, you would take the derivative of the area and set it equal to zero to find w.

  • Math -

    let the side parallel to the wall by y m
    let the other two sides be x m each

    so we have y + 2x = 24 or y = 24-2x
    area = xy
    = x(24-2x
    = -2x^2 + 24x
    let's complete the square to find the vertex of this parabola
    = -2(x^2 - 12x + 36 - 36)
    = -2( (x-6)^2 - 36)
    = -2(x-6)^2 + 72

    the vertex is (6,72)
    So the maximumum area is 72 ^2 , when x = 6 and y = 12

    or

    by calculus
    d(area) = -4x + 24 = 0 for a max area
    x = 6
    then y = 24 - 12 = 12
    max area = 6(12) = 72

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