math
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?

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 
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? 
Dick