Posted by Mary on Monday, March 5, 2012 at 7:52pm.
Normally 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.
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The 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 :)
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