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29 October, 13:02

Paul has 400 yards of fencing to enclose a rectangular area. Find the dimensions of the rectangle that maximize the enclosed area. What is the maximum area?

A rectangle that maximizes the enclosed area has a length of _ nothing yards and a width of _ nothing yards.

The maximum area is _ nothing square yards.

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  1. 29 October, 14:34
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    P=2 (x+y) perimeter equals two time the sum of x and y dimensions ...

    We are told that P=400 so

    2 (x+y) = 400

    x+y=200

    x=200-y ...

    Now area is:

    A=xy, using x found above we have:

    A = (200-y) y

    A=200y-y^2

    The rate of change of the area is dA/dy

    dA/dy=200-2y

    The maximum area occurs when the rate of change is zero or dA/dy=0 so

    200-2y=0

    2y=200

    y=x=100yd

    So the maximum area enclosed is when y=x=100 which is a perfect square (as is always the case for a four sided enclosure with a given amount of material)

    A=100^2=10000 yd^2
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