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

- precalculus -
**Anonymous**, Saturday, July 26, 2014 at 9:42pm
45000

## Answer this Question

## Related Questions

- Algebra 2 - A Rectangular Field is to be enclosed by a fence and divided into ...
- Algebra 2 - A Rectangular Field is to be enclosed by a fence and divided into ...
- math - A rectangular field is to be enclosed by a fence. Two fences parallel to ...
- Calculus - A rectangular field is to be enclosed by a fence and divided into ...
- Calculus 1 - A rectangular field is enclosed by a fence and seperated into two ...
- Math - A rectangular study area is to be enclosed by a fence and divided into ...
- calculus (optimization) - a rectangular study area is to be enclosed by a fence ...
- cal - a rancher has 4000 feet of fencing for constructing a rectangular corral. ...
- area and perimeter - a farmer wishes to build a fence around a rectangular field...
- math - Tshabalala wants to reconstruct his farm to separate his sheep and goats...

More Related Questions