Posted by sridhar on Wednesday, September 16, 2009 at 3:18pm.
You imposed two conditions on the problem
1. the length is twice the width, and
2. the sum of length and width is 15
so clearly 2x + x = 15
x = 5
So it no longer is a problem dealing with maximimums.
the enclosure is 5 by 10 for an area of 50
(the maximum area would have been obtained by having two equal sides of 7.5 feet for an area of 56.25 feet^2, clearly larger than the 50 from above. But your condition of length = twice the width would not have been followed)
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