Homework Help: precalculus

Posted by jeremy on Thursday, March 14, 2013 at 7:24am.

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.

precalculus - Reiny, Thursday, March 14, 2013 at 8:14am
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(600-2x)
= -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(x-150)^2 + 45000

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