We have 200 linear feet of fencing to enclose 2 adjacent rectangular corrals and we want to maximize the enclosed area. Let l be the length and w the width (see attachment) so:

`P=2l+3w` and`A=lw`

`200=2l+3w` ==>`2l=200-3w` ==>`l=100-3/2w`

`A=lw=w(100-3/2w)`

We are asked to maximize the area:

(1) Using algebra we see that A(w) is a parabola that opens down with intercepts at 0 and`200/3`. The vertex is at `w=100/3` so A=2172.2.

(2) Using calculus we find the derivative of A with respect to w to be 100-3w which is zero at `w=100/3` and the function has a maximum at`w=100/3`. The value of the function is A(100/3)=2172.2

The maximum area is 2172.2 sq ft and the dimensions are `w=100/3` ft and `l=50` ft

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