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 $y$?
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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$\begingroup$$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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