Friday

April 18, 2014

April 18, 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

**Related Questions**

Calc - A farmer wishes to enclose a rectangular pen with area 100 square feet ...

Calc - A farmer wishes to enclose a rectangular pen with area 100 square feet ...

math - a farmer has enough fencing to build 40 feet of fence. He wishes to build...

calculus optimization problem - A farmer has 460 feet of fencing with which to ...

math - A farmer will make a rectangular pen with 100 feet of fence using pasrt ...

calculus - a farmer has a 1500 feet of fencing in his barn.he wishes to enclose ...

calculus - Farmer Jones has 210 meters of fence. She wishes to construct a ...

Calculus 2 - A farmer wishes to build a fence for 6 adjacent rectangular pens. ...

math - A farmer has 1000ft of fence and wishes to enclose the largest possible ...

Mathematics - A farmer has both ducks and pigs in a pen together. There are a ...