Posted by awes on .
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,
length = x
width = y
area = A = x y
length of fencing = 2x+y = 40
so
y = 402x
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
(X10)^2 = (1/2)(A+200)
vertex at x = 10, A = 200 
math 
bobpursley,
40=2L+w
w=402L
area= LW=L(402L)
Using calculus
dArea/dL=0=402L + 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(402L)
graph area vs L. Where is it maximum? 
math 
Anonymous,
Dick