Posted by John on Friday, December 2, 2011 at 11:50pm.
A builder has 600 feet of fencing to enclose three adjacent rectangular partioned areas. Find the largest possible enclosed area of the partioned areas.

Precalculus  drwls, Saturday, December 3, 2011 at 12:38am
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

Precalculus (correction)  drwls, Saturday, December 3, 2011 at 1:43am
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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