Some details of the question appear to be missing.
I'm assuming the pen is to be rectangular and further that you want to find the length and width that enclose the maximum possible area.
Let the width of the pen be x, so that the length is 20-3x the area (y) is:
y = 20 x -2 x^2
y' = 20 -4 x
Maximum area exists when y'=0
Solve for x.
the challenge involves in dealing with the owner is to use a 20 meter fencing material to build a pig pen so that the owner will not be buying any additional materials considering also that one side of the pen is the outside wall of his garage? what is the dimension of this pig pen?
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