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November 26, 2014

November 26, 2014

Posted by **Shelby** on Tuesday, October 30, 2012 at 11:41pm.

How should the farmer allocate the fencing to the edges of the enclosure to maximize the area inside? That is, for the maximum area enclosure what should be the length of the side of the rectangle perpendicular to the barn and what should be the length of the side of the rectangle parallel to the barn?

- Calculus -
**Steve**, Wednesday, October 31, 2012 at 10:26amIf the length along the bar is x, and the width is y,

x+3y = 150

The area a = xy = (150-3y)y = 3(50y-y^2)

da/dy = 3(50-2y) da/dy=0 when y=25

thus, x = 75

So, the whole area is 75x25

As expected, the area is maximum when the fencing is evenly divided between lengths and widths. For a single fully enclosed area, that would be a square.

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