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March 28, 2015

March 28, 2015

Posted by **sridhar** on Wednesday, September 16, 2009 at 3:18pm.

- math -
**Reiny**, Wednesday, September 16, 2009 at 3:28pmYou 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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