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23 March, 06:56

You have a 64 foot of fencing. What are the dimensions of the rectangle of the greatest area you could enclose?

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  1. 23 March, 07:43
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    In a rectangle, opposite sides are congruent.

    Let one side have length x.

    The opposite side also has length x.

    The lengths of these two sides add to 2x.

    The lengths of all 4 sides add to 64, so the lengths of the other 2 sides

    add up to 64 - 2x. Each side measures 32 - x.

    The rectangle has sides of length x and 32 - x.

    The area of the rectangle is

    A = LW

    A = x (32 - x)

    A = 32x - x^2

    y = 32x - x^2 is a parabola that opens downward.

    The maximum value of the parabola is the vertex on top.

    32x - x^2 = 0

    (32 - x) x = 0

    32 - x = 0 or x = 0

    x = 32 or x = 0

    Since the parabola is symmetric with respect to the vertical axis, the vertex has x-coordinate 16.

    At x = 16, you get maximum area.

    Two opposite sides measure 16 ft each.

    32 - x = 32 - 16 = 16

    The other two opposite sides also measure 16 ft.

    Since all sides turned out to have length 16 ft, the rectangle is a square.

    Answer: The maximum area is enclosed by a square with side 16 ft.
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