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July 23, 2014

July 23, 2014

Posted by **km** on Thursday, June 28, 2012 at 9:39am.

- algebra -
**Steve**, Thursday, June 28, 2012 at 11:24amthe conditions require that the 4 posts are the corners of the rectangle. So, if the width is x, and the length is along the house, the fence used is x+x+(20-2x)

area = x(20-2x) = 2x(10-x)

the max is achieved when x is midway between the roots, or x=5.

The pen is thus 5x10, with area = 50

suppose the width is 6. The pen is then 6x8 with area = 48

suppose the width is 4. Then the pen is 4x12 with area = 48.

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