Problem source

$\,$
\begin{center}
\begin{tikzpicture}[scale=2]
    % Semicircle
    \def\r{2};
    \coordinate (D) at (0,0);
    \coordinate (A) at (-\r,0);
    \coordinate (B) at (0,\r);
    \coordinate (C) at (\r,0);
    \coordinate (X) at ({\r*cos(140)},{\r*sin(140)});
    \coordinate (Y) at ({\r*cos(75)},{\r*sin(75)});
    
    \draw (C) arc (0:180:\r);
    
    % Horizontal line
    \draw (A) -- (X) -- (Y) -- (C) -- (D) -- (Y) -- (X) -- (D) -- (A);
    
    % Labels
    \pic [draw, angle radius=1cm, "$\alpha$"] {angle = X--D--A};
    \pic [draw, angle radius=1.2cm, "$\theta$"] {angle = Y--D--X};
    \node at (Y) [right, above] {$Y$};
    \node at (X) [left] {$X$};
    \node at (B) [above] {$B$};
    \node at (A) [below, left] {$A$};
    \node at (D) [below] {$D$};
    \node at (C) [below, right] {$C$};
\end{tikzpicture}\end{center}
In the above diagram, $ABCD$ represents a semicircular window of fixed radius $r$ and centre $D$, and $AXYC$ is a quadrilateral blind. If $\angle XDY=\alpha$ is fixed and $\angle ADX=\theta$ is
variable, determine the value of $\theta$ which gives the blind $\textbf{maximum}$ area. 
If now $\alpha$ is allowed to vary but $r$ remains fixed, find the maximum possible area of the blind.

Solution source

The area for $\alpha$ fixed is $\frac12 r^2 \sin \alpha + \frac12 r^2 \sin \theta + \frac12 r^2 \sin (\pi - \theta - \alpha)$

So we wish to maximise $V = \sin \theta + \sin(\pi - \theta-\alpha)$

\begin{align*}
&& V &= \sin \theta + \sin(\pi - \theta-\alpha)\\
&&&= 2\sin \l \frac{\pi-\alpha}2\r\cos \l \frac{2\theta + \alpha - \pi}{2}\r
\end{align*}

The largest $\cos$ can be is $1$ when $\displaystyle 2\theta + \alpha - \pi = 0 \Rightarrow \theta = \frac{\pi-\alpha}2$. (ie we split the remaining area exactly in half).

We are now trying to maximise $W = \sin \alpha + 2 \sin \frac{\pi - \alpha}2$ ie

\begin{align*}
&& W &= \sin \alpha + 2 \cos \frac{\alpha}{2} \\
\Rightarrow && \frac{\d W}{\d \alpha} &= \cos \alpha-\sin \frac{\alpha}{2} \\
&&&= 1-2 \sin^2 \frac{\alpha}{2} - \sin \frac{\alpha}{2} \\
&&&= (1+\sin \frac{\alpha}{2})(1-2\sin \frac{\alpha}{2})
\end{align*}

Therefore $\frac{\alpha}{2} = -\frac{3\pi}{2}, \frac{\alpha}{2} = \frac{\pi}{6}, \frac{5\pi}{6} \Rightarrow \alpha = -3\pi, \frac{\pi}{3}, \frac{5\pi}{3}$. The only turning point in our range is $\frac{\pi}{3}$. This is obvious a a maximum by symmetry or checking the end points, but we could also check the second derivative $\frac{\d^2 W}{\d \alpha^2} = -\sin \frac{\pi}{3}-\cos \frac{\pi}{3} < 0$ so we have a maximum.

Therefore the largest possible area is: $\displaystyle \frac{3\sqrt{3}}{4}r^2$