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 +...
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