<PRE><B>Question 134979</B>
A rectangular playground is to be fenced off and divided in two by another fence parallel to one side of the playground. 696 feet of fencing is used. Find the maximum area of the playground.
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Important: Draw the figure so you can follow this solution.
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There are 2 length lines and 3 width lines
2L + 3W = 696 ft
L = (-3/2)W+ 348
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Area = LW
Area = W[(-3/2)W+348]
Area = (-3/2)W^2 + 348W
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Comment: This is a quadratic with a = -3/2 and b = 348
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Maximum area occurs at w = -b/2a = -348)/(-3) = 116
The maximum area is (-3/2)(116)^2+348(116) = 20184 sq. ft.
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Cheers,
Stan H.</PRE>