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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*Sin%5Bx%5D%2C%7B‌​x%2C0%2CPi%2F2%7D%5D) – gt6989b Dec 19 '12 at 16:40
why would they put this on a practice test of for the ap bc exam? – yiyi Dec 19 '12 at 16:45
@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... – gt6989b Dec 19 '12 at 16:49
From here: "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.) – David Mitra Dec 19 '12 at 16:49
@DavidMitra isn't that what I did? – yiyi Dec 19 '12 at 16:58
up vote 2 down vote accepted

$A'(x)=0$ 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

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