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Calculus/Calculus Applications of Differentiation
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Question
a rancher has 400 feet of fencing with which to enclose two adjacent rectangular corrals. What dimensions should be used so that the enclosed area will be maximum?
Make a rectangle out of the fence that is A x B.
Cut it in half from the middle of each side of length B, giving another fence of length A.
We now have 2B + 3A as the total fencing with A/2 as the width of each corral.
Since the total fencing is 400', we know that 2B + 3A = 400, or 2B = 400-3A, so B = (400-3A)/2.
The area of both of the pastures would be AB, and that is A(400-3A)/2.
Multiplying it out gives 20A - 3Ał/2, and that I will call f(A).
Now f'(A) = 200 - 3A, and setting this to 0 will give us where the maximum is,
since f(A) is a downward opening parabola.
For 200 - 3A = 0, we need 200 = 3A, which means that A = 200/3.
Since 3A = 200, that means that 2B = 400 - 3A can be rewritten as 2B = 400 - 200 = 200.
This means that B = 200/2 = 100.
Thus, for the adjacent equal rectangular corrals, the width of each would be B/2 = 50
and the length A would be 200/3. That makes for an area inside of each as 10,000/3.
Since they both have equal size, the area of both is 20,000/3 or 6,666 2/3.
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