Following the suggestion of Andreas Blass in his comment, I'll look at the question of the varying area of the axis-alligned "bounding box" rectangle $B$ as the rectangle $A$ rotates.
Suppose $A$ has side lengths $a,b$. Put one vertex at the origin, and another at $P=(a \cos t,a \sin t)$. The vertex making the right angle at $(0,0)$ with $P$ is then at $Q=(-b \sin t, b \cos t).$ The fourth vertex, diagonally opposite $(0,0)$, ends up (using vector addition) at
$$R=(a \cos t-b \sin t, a \sin t + b \cos t).$$
Now for $0 \le t \le \pi/2$ it's clear that the horizontal width of the bounding box $B$ occurs because of the points $P,Q$, i.e. the left side of the bounding box goes through $Q$ while its right side goes through $P$. Thus the width of $B$ is the difference of $x$ coordinates of $P,Q$, i.e. $a \cos t + b \sin t$. At the same time the height is due to the origin and the point $R$ and is the difference of their $y$ coordinates, i.e. the height of $B$ is $a \sin t + b \cos t$.
This gives the area $A(t)$ of the bounding box, in terms of $t$ with $0 \le t \le \pi/2$, as
$$A(t)=( a \cos t + b \sin t)(a \sin t + b \cos t).$$ After applying pythagoras' identity and a double angle formula we can rewrite $A(t)$ in the convenient form
This makes it clear that the maximum area of $B$ occurs when $t=\pi/4$ where the sine term is $1$, making the maximal area
Side note: the bounding rectangle turns out to be a square at angle 45 degrees from the $x$ axis, with the rotating rectangle $A$ going along diagonal of the bounding box (bounding square). That makes me think there is some kind of symmetry here which would make a more geometric argument possible.