Posted by **awes** on Tuesday, September 13, 2011 at 4:37am.

a farmer has enough fencing to build 40 feet of fence. He wishes to build a rectangular pen nest to his barn wall forming one side of the pen. What dimension should he make the pen so as to enclose the greatest possibility area?

- math -
**Damon**, Tuesday, September 13, 2011 at 5:35am
length = x

width = y

area = A = x y

length of fencing = 2x+y = 40

so

y = 40-2x

A = x (40 - 2x)

A = 40 x - 2 x^2

maxmimum A , when dA/dx = 0

0 = 40 - 4 x

x = 10

then y = 40 - 20 = 20

area = A = 200

alternatively look at parabola

2 x^2 -40 x = -A

x^2 - 20 x = -(A/2)

x^2 - 20 x + 100 = -(A/2) + 100

(X-10)^2 = -(1/2)(A+200)

vertex at x = 10, A = 200

- math -
**bobpursley**, Tuesday, September 13, 2011 at 5:38am
40=2L+w

w=40-2L

area= LW=L(40-2L)

Using calculus

dArea/dL=0=40-2L + l(-2)

4L=40

L=10 W= 20 so the pen is 10x20,with the barn serving as a 20ft side.

non calculus?

area=L(40-2L)

graph area vs L. Where is it maximum?

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