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
 A: 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 $[0,\pi/2]$.


A: 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.
