# Trigonometry Min/Max Problem

$f(x) = 2\sin x \hspace{10pt}(0 \leq x \leq \pi)$
$g(x) = -\sin x \hspace{10pt}(0 \leq x \leq \pi)$

Rectangle ABCD is enclosed between the above functions' graphs (its edges are parallel to the axes).

How would I go about finding the maximum perimeter of ABCD?

I'm really clueless about this, I don't even know how to begin. How am I supposed to represent the edges?

Thanks

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If the left vertical side of the rectangle passes through $x=a$ then the height of the rectangle is $2\sin a +\sin a=3\sin a$. The right vertical side of the rectangle then passes through $\pi-a$ (draw a picture). So the width of the rectangle whose left vertical side passes through $x=a$ is $(\pi-2a)$. The perimeter of the rectangle is then $$2\cdot 3\sin a +2(\pi-2a).$$
You need to find the maximum value of the above expression over the interval $[0,\pi/2]$.

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Great, thanks!! – Lior Mar 16 '12 at 13:33
@Lior You're welcome. I misread your post though, and just noted you want to maximize the perimeter. I originally had the area, but edited accordingly. – David Mitra Mar 16 '12 at 13:45

As a minor matter of strategy, after drawing the picture, I would move the $y$-axis to get symmetry. Then the picture becomes the picture of $2\cos x$ and $-\cos x$. If the upper right-hand corner of the rectangle in the new picture is $(x,2\cos x)$, then the perimeter is $4x+6\cos x$.

We have $0\le x\le \pi/2$, and the maximum is obviously not reached at an endpoint. So the maximum is reached where the derivative of $4x+6\cos x$ is $0$. This happens where $4-6\sin x=0$, that is, where $\sin x=\frac{2}{3}$. There, $x=\arcsin(2/3)$ and $\cos x=\frac{\sqrt{5}}{3}$, and now we can calculate the maximum area.

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