calculus
posted by lisa .
Suppose you have 400 meters of fencing and you wish to enclose a rectangular space with two parallel partitions: What is the maximum total area of the enclosed space, in square meters?

If the original rectangle is x m wide and y m long
then the partitions are x m as well
so we have
2y + 4x = 400
y + 2x = 200
y = 200  2x
Area = xy
A = x(2002x) = 200x  2x^2
dA/dx = 200  4x = 0 for a max of A
200 = 4x
x = 50
then y = 200  100 = 100
if x = 50 and y = 100, we get a maximum area
which is xy or 50(100) = 5000 m^2
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