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November 27, 2014

November 27, 2014

Posted by **Anonymous** on Tuesday, October 25, 2011 at 2:48pm.

- Calculus -
**Steve**, Tuesday, October 25, 2011 at 3:24pmIf each pen has width x and length y (against the barn),

the area is 4xy, and the fence used is 5x+8y

So, each pen has area 100.

p = 5x + 8(100/x) = 5x + 800/x

dp/dx = 5 - 800/x^2

dp/dx = 0 when x = √160 = 4√10

x = 4√10

y = 25/√10

- Calculus -
**Steve**, Tuesday, October 25, 2011 at 3:27pmNow, if the barn is used as one wall of the pen, meaning only 3 sides have to be fenced, then

p = 5x+4y = 5x + 4(100/x) = 5x + 400/x

p' = 5 - 400/x^2

p' = 0 at x = 4√5

y = 25/√5

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