Precalculus
posted by John .
A builder has 600 feet of fencing to enclose three adjacent rectangular partioned areas. Find the largest possible enclosed area of the partioned areas.

I will assume the maximum area is achieved with three adjacent rectangular areas with total length (end to end) = y and width = x.
Total area = x * y
2y + 3x = 600
A = x * (300  1.5 x) = 300 x 1.5 x^2
dA/dx = 0
300 = 3x
x = 100
y = (600  300)/2 = 150
Total area = x*y = 15,000 
I just realized that the equation that relates x and y to the total length of material should be
2y + 4x = 600.
You need 4 fence segments of length x to create 3 areas.
Therefore, y = 300  2x
A = x*(300  2x) = 300x  2x^2
Since dA/dx = 0,
4x = 300 and x = 75.
y = 150
Maximum total area = x*y = 11,250
You say you are takng a precalculus course, but I have used calculus in this derivation.
The same result can be obtained by "completing the square" with the A(x) function.
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