precalculus
posted by jeremy .
A rectangular field is to be enclosed by a fence and divided into three equal rectangular parts by two other fences. find the maximum area that can be enclosed and separated in this way with 1200m of fencing.

let the width of the whole rectangle be x m (there will be 4 of these)
let the length be y m
then 4x + 2y = 1200
2x + y = 600
y = 600  2x
Area = xy
= x(6002x)
= 2x^2 + 600x
Now, I don't know if you are studying Calculus.
If you do, then
d(Area)/dx = 4x + 600
= 0 for a max area
x = 150
then y = 600  2(150) = 300
and the max area is (150)(300) = 45000
If you don't know Calculus, complete the square on the above quadratic
you should end up with
Area = 2(x150)^2 + 45000 
45000
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