Length and breadth of a rectangle enclosed between two semi-circles of given radii 
34. It is required to take a rectangular frame in a horizontal position along a corridor bounded by vertical walls of which a horizontal cross-section is two concentric semicircles of radii $r$ and $r\ \sqrt{3}$; the frame is of length $2x$ and breadth $y$. One side of length $2x$ is tangential to the inner wall, and the two ends of the opposite side are in contact with the outer wall, as shown in Fig. Misc. Ex. 1-34. Prove that
$$x^{2} = 2r^{2} - 2ry - y^{2}.$$
Prove that, if $x$ and $y$ may vary, the greatest possible area enclosed by the frame is $\frac{1}{2}r^{2}\ \sqrt{3}.$


I noticed a few things that I believe are relevant. The first is that one of the sides of the rectangle is a chord of the larger circle and the second is that the angle between the radius of the smaller circle and the opposite side of rectangle is 90 degrees.
It's been a long time since I had fun with circles, but using Wolfram MathWorld I was able to piece together this much:

$2x = 2r\sqrt{3}\sin\left(\dfrac{\theta}{2}\right) \Leftrightarrow x = r\sqrt{3}\sin\left(\dfrac{\theta}{2}\right)$
$h = r\sqrt{3} - (y + r) \Leftrightarrow y = r\sqrt{3} - r - h$
$y + r = r\sqrt{3}\cos\left(\dfrac{\theta}{2}\right) \Leftrightarrow y = r\sqrt{3}\cos\left(\dfrac{\theta}{2}\right) - r$
$x^{2} = 3r^{2}\sin^{2}\left(\dfrac{\theta}{2}\right)$
$y^{2} = 3r^{2}\cos^{2}\left(\dfrac{\theta}{2}\right) - 2r^{2}\sqrt{3}\cos\left(\dfrac{\theta}{2}\right) + r^{2}$
But that's about everything of use - does anybody have any hints?
Edit to incorporate the suggestion of André Nicolas:
May we form a triangle as shown in dashed pink and so derive $y = \sqrt{3r^{2} - x^{2}} - r$

 A: A start: By the Pythagorean Theorem we have $y=\sqrt{3r^2-x^2}-r$, which can be rewritten as $y+r=\sqrt{3r^2-x^2}$. Square both sides and simplify.
For the area maximization, equivalently maximize $x^2y^2$, that is, $2r^2y^2-2ry^3-y^4$.
A: Place the circle centers at the origin of a Cartesian grid. Without loss of generality, we can assume optimal rectangle is oriented with its base parallel to the $X$ axis, touching the smaller circle at the base midpoint at $(r,0)$.
Then the coordinates of the upper right hand corner can be taken to be $(rp,rq)$ where, since it must be on the outer circle, 
$$
q = \sqrt{3-p^2}$$
Then the width of the rectagle is $2rp$ and the height is $r(q-1) = r(\sqrt{3-p^2}-1)$.
The area is 
$$A=wh = 2r^2p(\sqrt{3-p^2}-1)$$
$$\frac{dA}{dp} = 2r^2\left[\sqrt{3-p^2}-1-\frac{p^2}{\sqrt{3-p^2}} \right]$$
We can set this to zero and solve for $p$ but since this is not easy I will show the steps. The expression is simpler if we solve for $q$, noting that $p^2 = 3-q^2$.
$$ q-1- \frac{3-q^2}{q}=0\\
q^2 - q - 3 + q^2 = 0 \\
2q^2 -q+3 = 0 \\
q = \frac{1+\sqrt{1+4\cdot 3\cdot 2}}{4}= \frac{3}{2}
\\p = \frac{\sqrt{3}}{2}
\\A_{\max} = 2r^2p(q-1) = 2r^2\frac{\sqrt{3}}{2}\frac{1}{2}=\frac{\sqrt{3}r^2}{2}
$$
