Posted by anonymous on Wednesday, December 9, 2009 at 11:22pm.
Let the combined length of the two rectangles by y
let the width be x (there will be 3 of those)
so 3x + y = 132
y = 132-3x
Area = xy
= x(132-x)
= -x^2 + 132x
by Calculus
d(Area)/dx = -2x + 132 = 0 for a max of Area
x = 66
then max Area = 66(132-66) = 4356 m^2
by completing the square:
Area = -[x^2 - 132x + 4356 - 4356 ]
= -(x-66)^2 + 4356
so the max Area is 4356 , when x = 66
thanks!
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