Saturday

August 30, 2014

August 30, 2014

Posted by **sylvia** on Saturday, November 15, 2008 at 11:02am.

- calculus -
**drwls**, Saturday, November 15, 2008 at 11:45amThe curve describes the arch of a cosine function that goes from 0 at x = pi, to 1 at x = 0, and back down to 0 at x = pi. Pick two points on that curve, at x = + and - a, to construct an inscribed rectangle. The dimensions of the rectangle will be width = 2a and height = 4 cos (a/2).

The area will be A(a)= 8 a cos (a/2).

Find the value of a that maximizes this area

dA/da = 0

8 cos (a/2) -8a sin (a/2) = 0

a = cot (a/2)

That will have to be solved by iteration or graphing. I get a = 1.306

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