Posted by anonymous on Wednesday, December 9, 2009 at 11:22pm.
How should this be done?
Suppose you have 132 m of fencing with which to make two sidebyside rectangular enclosures against an existing wall. if the rectangular enclosures are adjacent and of the same depth what is the maximum are that can be enclosed?

MATH ANALYSIS  Reiny, Wednesday, December 9, 2009 at 11:46pm
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 = 1323x
Area = xy
= x(132x)
= x^2 + 132x
by Calculus
d(Area)/dx = 2x + 132 = 0 for a max of Area
x = 66
then max Area = 66(13266) = 4356 m^2
by completing the square:
Area = [x^2  132x + 4356  4356 ]
= (x66)^2 + 4356
so the max Area is 4356 , when x = 66

MATH ANALYSIS  anonymous, Wednesday, December 9, 2009 at 11:50pm
thanks!
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