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 side-by-side 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 = 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

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

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