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29 September, 21:17

the city park department is planning to enclose a play aea. one side of the area will be against an existing building, so no fence is needed there. find the dimensions ofthe maximum rectangular area that can be enclosed with 800m of fence

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  1. 29 September, 21:37
    0
    length = 200 m

    width = 400 m

    Step-by-step explanation:

    Let the length of the plaing area is L and the width of the playing area is W.

    Length of fencing around three sides = 2 L + W = 800

    W = 800 - 2L ... (1)

    Let A is the area of playing area

    A = L x W

    A = L (800 - 2L)

    A = 800 L - 2L²

    Differentiate with respect to L.

    dA/dL = 800 - 4 L

    It is equal to zero for maxima and minima

    800 - 4 L = 0

    L = 200 m

    W = 800 - 2 x 200 = 400 m

    So, the area is maximum if the length is 200 m and the width is 400 m.
  2. 29 September, 21:45
    0
    Dimensions : Length = 400, Width = 200

    Step-by-step explanation:

    Let length & width of rectangle park be = L, W. Also, let one side of length be supported by building wall, so not needing fencing.

    So, the perimeter of rectangle park, including 2 width & 1 length:

    2W + L = 800

    L = 800 - 2W

    Rectangle Area = Length x Width

    A = L x W

    A = (800 - 2W) W

    A = 800W - 2W^2

    To maximise area, it will have to be differentiated w. r. t dimension width

    dA / dW = 800 - 4W

    dA / dW = 800 - 4W = 0

    800 = 4W

    W = 800 / 4

    Width [W] = 200

    Length [L] = 800 - 2W

    = 800 - 2 (200)

    = 800 - 400

    Length [L] = 400
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