Posted by pam on Thursday, January 8, 2009 at 4:12am.
Do 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.
I 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
I misread the problem, and forgot that the fence had to also enclose the outer perimter, which was clearly stated.
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