# math

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?

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1. 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

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2. 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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bobpursley

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