Two adjacent rectangular corrals are to be enclosed with 200 feet of fencing. Let us assume that the dimensions of the two rectangles are the same.

If the dimensions are w and l, the total perimeter of the two rectangular corrals is 4w+4l but with a common side this is equal to 4w + 3l.

Now the enclosed area is 2*w*l and as the length of the fencing is 200 ft, we have l + 2w + 2w + l + l = 200

3l + 4w = 200

`l = (200 - 4w)/3`

To maximize the enclosed area, we need to maximize: `A = 2*w*l = 2*w*(200 - 4w)/3`

`A = (400w - 8w^2)/3`

Using calculus, equate A' = 0 and solve for w:

`A' = ((400w - 8w^2)/3)'`

`= 400/3 - (16/3)*w`

`400/3 - (16/3)*w = 0`

`=> 400 - 16w = 0`

`=> w = 400/16`

`=> w = 25`

`l = (200 - 4w)/3 = 100/3`

**The maximum enclosed area is 5000/3 square feet.**

Denote the length of the shared side of the corrals by `l` and the width of the both corrals together as `w` (please see the attached image). Then, the required amount of fencing will be

`P = l +...

(The entire section contains 3 answers and 519 words.)

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Here is a similar problem with 400 total feet of fencing instead of 200.