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November 28, 2014

November 28, 2014

Posted by **pam** on Thursday, January 8, 2009 at 4:12am.

- calculus -
**drwls**, Thursday, January 8, 2009 at 5:57amDo you mean to minimize the amount of FENCE used? I don't see what force has to do with it. It also seeme to me that, by making the outer rectangle very narrow, the amount of fence required to divide it in half can be reduced to zero.

- calculus -
**Damon**, Thursday, January 8, 2009 at 6:00amI assume you mean "fence" used.

Three cross fences of length w

area = w L = 384

so w = 384/L

total fence length= t = 3w + 2 L

t = 3*384/L + 2 L

t = 1152/L + 2 L

for min dt/dL = 0

dt/dL = 0 = -1152/L^2 + 2

2 L^2 = 1152

L^2 = 576

L = 24

then w = 16

- calculus -
**drwls**, Thursday, January 8, 2009 at 6:16amI misread the problem, and forgot that the fence had to also enclose the outer perimter, which was clearly stated.

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