First of all, there may be multiple rectangles that satisfy your conditions, e.g.
if you want specific angle of rectangle, or
if you want to specify a point on the boundary.
However, there is a special case where there is at most only once rectangle, i.e. if you assume that it has to touch both of your boundaries. In such case it is easy to compute it. Place the origin (point $(0,0)$) in the center of the boundary (the center of the circle), and let $(C_x,C_y)$ be the top right boundary corner (the boundary rectangle has size $2C_x \times 2C_y$ ). Let $(x,y)$ be the top right vertex of small rectangle, then it satisfies conditions ($\alpha > 0$ means counterclockwise rotation):
\begin{align*}
\left[\begin{matrix}x'\\\ C_y\end{matrix}\right] &= \left[\begin{matrix}\cos\alpha&-\sin\alpha\\\sin\alpha&\cos\alpha\end{matrix}\right]\left[\begin{matrix}x\\\y\end{matrix}\right] \\\
\left[\begin{matrix}C_x\\\ y'\end{matrix}\right] &= \left[\begin{matrix}\cos\alpha&-\sin\alpha\\\sin\alpha&\cos\alpha\end{matrix}\right]\left[\begin{matrix}x\\\ -y\end{matrix}\right]
\end{align*}
where $x'$ and $y'$ are just placeholders. Extracting appropriate rows from those formulae, we can transform that into one equation (notice the lack of minus sign in the matrix):
\begin{align*}
\left[\begin{matrix}C_x\\\ C_y\end{matrix}\right] &= \left[\begin{matrix}\cos\alpha&\sin\alpha\\\sin\alpha&\cos\alpha\end{matrix}\right]\left[\begin{matrix}x\\\ y\end{matrix}\right]
\end{align*}
with solution being:
\begin{align*}
\left[\begin{matrix}\cos\alpha&\sin\alpha\\\sin\alpha&\cos\alpha\end{matrix}\right]^{-1}
\left[\begin{matrix}C_x\\\ C_y\end{matrix}\right] &= \left[\begin{matrix}x\\\ y\end{matrix}\right] \\\
\sec{2\alpha}\left[\begin{matrix}\cos\alpha&-\sin\alpha\\\ -\sin\alpha&\cos\alpha\end{matrix}\right]
\left[\begin{matrix}C_x\\\ C_y\end{matrix}\right] &= \left[\begin{matrix}x\\\ y\end{matrix}\right]
\end{align*}
Please note, that this may not have a proper solution if $\alpha$ is to big!
Edit: Ok, I missed the comment about maximizing the area. Then again, consider this example:
The gray figure is a rhombus (it was created by rotating the black rectangle by $2\alpha$ ). To get the inscribed rectangle with the greatest area, consider the case when the rhombus would be a square--the greatest area would be when each rectangle vertex splits the rhombus edge in half (because then it is also a square and that is the rectangle with greatest area and given perimeter).
But we can scale our rhombus (that may not be a square) so that is a square, obtain the solution there, and then scale back (the area will scale accordingly)! In conclusion the rectangle of greatest area will split the rhombus edges in half.
How to compute it? You could do it using the same approach:
\begin{align*}
\left[\begin{matrix}x'\\\ C_y\end{matrix}\right] &= \left[\begin{matrix}\cos\alpha&-\sin\alpha\\\sin\alpha&\cos\alpha\end{matrix}\right]\left[\begin{matrix}x\\\y\end{matrix}\right] \\\
\left[\begin{matrix}x''\\\ -C_y\end{matrix}\right] &=
\left[\begin{matrix}\cos(-\alpha)&-\sin(-\alpha)\\\sin(-\alpha)&\cos(-\alpha)\end{matrix}\right]
\left[\begin{matrix}x\\\ -y\end{matrix}\right]
\end{align*}
However, one can do it simpler: the vertex of the inscribed rectangle splits the
gray edge in half, so $$2x\sin\alpha = C_y = 2y\cos\alpha\,.$$ This works if $C_x > C_y$, otherwise you need to do the same for $C_x$ instead. Moreover, even if $C_x > C_y$, you still need to check if the rotated rectangle fits into the boundary (because it may be that $C_x = C_y + \varepsilon $ ), if not, the solution from previous part will do.
Hope that helps ;-)
Best Answer
Say the length and width are $L,W$. We are given values $A,B$ with $$\frac LW = A\quad\&\quad L^2+W^2=B^2$$
We note that $$L=AW\implies (1+A^2)W^2=B^2\implies W=\frac B{\sqrt {1+A^2}}$$
It follows that $$Area =LW = AW^2= \frac {AB^2}{1+A^2}$$