Answer:
- overall: 50 ft by 33 1/3 ft
- each pen: 25 ft by 33 1/3 ft
Explanation:
You want the dimensions of the maximum rectangular area that can be enclosed by 200 feet of fencing if there is one partition in the middle of it.
Area
We assume the identical adjacent rectangular corrals can be constructed by putting a partition down the middle of a larger rectangular corral. If the overall length of that larger corral is x, then the area will be ...
total fence used on long sides: 2x
total fence used on short sides and partition: 200 -2x
length of short side: (200 -2x)/3
overall area: 2x(200 -2x)/3 = (4/3)(100x -x²)
Length
The length is the value of x that makes the area derivative zero:
dA/dx = (4/3)(100 -2x) = 0
100 = 2x . . . . . . make second factor zero
50 = x
The length of the short side is then ...
(200 -2x)/3 = (200 -100)/3 = 33 1/3
The enclosed area should be 50 ft long by 33 1/3 ft wide, with a 33 1/3 ft partition in the middle of the long side. Each pen will be 33 1/3 ft by 25 ft.
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Additional comment
This problem can be solved without calculus by recognizing the area function as the equation of a parabola that opens downward. It has zeros at x=0 and x=100, so a maximum at its vertex at x=50.
The generic solution to a problem like this one is that the long sides use half the fencing and the remaining sides use half the fencing. Here, the two long sides total 200 ft/2 = 100 ft, so are 50 ft each. The remaining three lengths of fence total the other half, 100 ft, so the ends and center partition are 100/3 = 33 1/3 ft.
Often, a barn or river is involved, so that the fencing only needs to be used for 3 sides, The division of fence is the same. For maximum area, half is used in one direction, and the other half is used in the orthogonal direction.
If the cost of fence is different for each side or partition, the cost is minimum when it is split evenly between the orthogonal directions. The cost of the east-west sides will equal the cost of the north-south sides.