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December 18, 2014

December 18, 2014

Posted by **Mary** on Monday, March 5, 2012 at 7:52pm.

The maximum area that the farmer can enclose with 80 ft of fence is ___sq ft.

The dimensions of the garden to give this area is 40 ft by ____ft

- algebra -
**tom**, Monday, May 14, 2012 at 11:05amNormally the largest area is enclosed by a square but this is only when the perimeter is 4 sides. In this problem the barn is the forth side with the 80 ft of fence used for only three sides. The complication is that the barn can contribute any lenght for its side. So the barn length as a variable turns the problem into a second order equation or the quadratic.

- algebra -
**Anonymous**, Wednesday, June 20, 2012 at 8:32pmc15678e

- algebra -
**Lucas**, Tuesday, July 24, 2012 at 9:18amThe equations are 2x+y=80 and 2x=y. "x" being the widthof the fence coming off the barn and "y" being the length connecting those two pieces. When one side is as big as nessasary (aka the barn, river, wall, etc.) the length is double the width. The perimeter is usually defined =2x+2y, but in this case one y is already there so the equation 2x+y=80 (and 2x+y=40)works. Try that out :)

- algebra -
**Anonymous**, Thursday, December 11, 2014 at 11:26pmhjhj

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