# Math

Find the maximum area of a rectangle inscribed in the region bounded by the graph of
y = (3-x)/(2+x)

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1. since the center of the rectangle will lie at (-2,-1) if it extends x units right and left of x=-2, then the area of the upper-right quarter is x*5/x = 5

So, the rectangle has a constant area of 5!

If you don't like that answer, obtained by shifting the graph right 2 and up 1 so that it becomes just y=5/x then plug in the real x-y values. If the upper right coordinate of the square is (x,y) then the upper-right area is
(x+2)[((3-x))/((x+2))+1]
= (x+2)(3-x+x+2)/(x+2)
= (x+2)(5)/(x+2)
= 5

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