A rancher has 200 feet of fencing to enclose two adjacent rectangular corrals, what dimension should be used to so that the enclosed area will be a maximum?

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 andA=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 and200/3. 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 atw=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