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 = 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, Tuesday, September 13, 2011 at 5:38am
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, Sunday, November 13, 2016 at 11:26am
Dick
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