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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