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A rectangle with one side on the $x$-axis has its upper vertices on the graph of $y = \cos x$. What is the minimum area outside the rectangle but under the graph of $y$?

enter image description here

My work so far

I was thinking of just dealing with $[0,\frac{\pi}{2}]$

Then I would need to just deal with one vertice. Then, let $A\left(x\right)$ be the area of the rectangle inside. When is $A\left(x\right)$ largest?

I think that $A\left(x\right) =x * \cos(x)$ length*height

$A^{\prime}\left(x\right) = \cos\left(x\right) - x\sin\left(x\right)$

How to find the values when $A^{\prime}\left(x\right)=0$ Thanks gt6989b

So $\approx 0.860$

Thus the shaded area will be $2[1-0.860\cos\left(0.860\right)]\approx 0.877$

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    The equation $x = \cot x$ is transcendental, the only root in $[0,\pi/2]$ is approximately $0.86$, as computed by Mathematica (see http://www.wolframalpha.com/input/?i=NSolve%5BCos%5Bx%5D%3D%3Dx*Sin%5Bx%5D%2C%7Bx%2C0%2CPi%2F2%7D%5D)2012-12-19
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    why would they put this on a practice test of for the ap bc exam?2012-12-19
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    @anonymous not quite - remember, it's only being considered on $[0, \pi/2$, not on $[-\pi/2,\pi/2]$ as is plotted in the picture...2012-12-19
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    From [here](http://apcentral.collegeboard.com/apc/members/exam/exam_information/8031.html): "Because graphing calculator use is an integral part of the course, the exam contains questions that require students to use a graphing calculator.". (Which is somewhat sad, in my opinion. You could simply graph $y=A(x)$ and find the maximum value of $A$ from that.)2012-12-19
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    @DavidMitra isn't that what I did?2012-12-19
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    I just meant to suggest that you can avoid the derivative analysis if you're allowed to solve the problem "graphically".2012-12-19

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$A'(x)=0$ is equivalent to $x=\cot(x)$ or $x=1 / \tan(x)$.

There are several solutions, but the only one in $[0,\frac{\pi}{2})$ is about $0.860333589$

I suspect there is no closed form, but it is easy to find a close enough value by numerical approximation.

You then need to multiply this by its cosine, double it, and subtract the result from the total area under the curve