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December 19, 2014

December 19, 2014

Posted by **Andres** on Wednesday, October 12, 2011 at 8:04pm.

- calc -
**bobpursley**, Wednesday, October 12, 2011 at 8:14pmArea= R*W where road fence is R, W is the length perpendicular to the Road.

Cost= 34R+16(R+2W)= 34Area/W+ 16(Area/W+2W)

dCost/dw= -34*100/W^2+ 16(-100/W^2+2)=0

0=-3400-1600+32W^2

W= sqrt (5000/32)

L= 100/W

check my math.

- calc -
**Steve**, Wednesday, October 12, 2011 at 8:16pmlet there be length a and width b, with side a along the road.

b = 100/a

cost is a*34 + a*16 + 2*100/a * 16

c = 50a + 200/a

c' = 50 - 200/a^2

c' = 0 when a = 2

so, the minimum cost is 100 + 100 = 200

a 2' wide pen? Is he housing gerbils?

- calc -
**Steve**, Thursday, October 13, 2011 at 12:07amMy bad - bobpursley is correct. The road length is 8, width is 12.5

using my notation,

c = 50a + 3200/a

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