# Fence Max Area Problem -- am I right??

• Dec 3rd 2009, 11:45 PM
golfman44
Fence Max Area Problem -- am I right??
Q: a farmer has 4800m of fencing with which to build a rectangular pen. The pen is to be divided into 3 euqal parts by fences parallel to the sides of the pen. The farmer wishes to maximize the total area of these pens. What will be the maximum area under these conditions?

Drawing is just a wide rectangle split into three equal parts....Y is the top and bottom, X are the four sides

A = xy
4x+2y = 4800

y = (4800-4x)/2 --> y = 2400 - 2x

A = x(2400 - 2x) --> A = 2400x - 2x^2

A(x) = 2400x - 2x^2
A'(x) = 2400 - 4x --> x = 600 @ a = 0

X = 600
Y = 1200

A = 720,000 M^2
• Dec 3rd 2009, 11:56 PM
dedust
in the question, total length of the fence = 400, but you wrote 4x+2y = 4800, typo??

if total length of the fence = 4800, you are doing right (Clapping)(Clapping)(Clapping)
• Dec 4th 2009, 12:00 AM
golfman44
Quote:

Originally Posted by dedust
in the question, total length of the fence = 400, but you wrote 4x+2y = 4800, typo??

if total length of the fence = 4800, you are doing right (Clapping)(Clapping)(Clapping)

Yes, that was a typo. It is indeed 4800. So the answer is correct??
• Dec 4th 2009, 12:30 AM
dedust
yes mate, you're correct