Problem source

Sketch the curve $y^{2}=1-\left|x\right|$. A rectangle, with sides parallel to the axes, is inscribed within this curve. Show that the largest possible area of the rectangle is $8/\sqrt{27}$. 
Find the maximum area of a rectangle similarly inscribed within the curve given by $y^{2m}=\left(1-\left|x\right|\right)^{n}$, where $m$ and $n$ are positive integers, with $n$ odd.

Solution source

\begin{center}
    \begin{tikzpicture}[scale=2]
        \draw[->] (-1.25, 0) -- (1.25, 0);
        \draw[->] (0, -1.25) -- (0, 1.25);

        \node at (1.25, 0) [right] {$x$};
        \node at (0, 1.25) [above] {$y$};

        \node at (1.3, 0.75) {$y^2 = 1- |x|$};

        \draw[domain = -1:1, samples=200, variable = \x]  plot ({\x},{sqrt(1-abs(\x))});
        \draw[domain = -1:1, samples=200, variable = \x]  plot ({\x},{-sqrt(1-abs(\x))});

        \draw ({1-1/100}, {1/10}) -- (1, 0);
        \draw ({1-1/100}, {-1/10}) -- (1, 0);
        
    \end{tikzpicture}
\end{center}

Suppose one of the $x$ coordinates is $t > 0$, then the coordinates are $y = \pm \sqrt{1-t}, x = \pm t$. The area will be $A = 2t \cdot 2 \sqrt{1-t}$. 

To maximise this,

\begin{align*}
&& \frac{\d A}{\d t} &= 4 \sqrt{1-t} - 2t(1-t)^{-\frac12} \\
&&&= \frac{4(1-t) - 2t}{\sqrt{1-t}} \\
&&&= \frac{4-6t}{\sqrt{1-t}}
\end{align*}

Therefore there is a stationary point at $t = \frac23$. Since we know the area is $0$ when $t = 0, 1$ we can see this must be a maximum for the area.

Therefore the area is $\displaystyle 4 \frac23 \sqrt{1-\frac23} = \frac{8}{3\sqrt{3}} = \frac{8}{\sqrt{27}}$.

For this similar problem, using a similar approach we find $y = \pm (1- t)^{n/2m}, x = \pm t$ and so the area is $A = 4 t \cdot (1-t)^{n/2m}$.

\begin{align*}
&& \frac{\d A}{\d t} &= 4(1-t)^{n/2m} - 4t \frac{n}{2m} (1-t)^{\frac{n}{2m} - 1} \\
&&&= (1-t)^{\frac{n}{2m}-1} \left ( 4(1-t) - \frac{2n}{m} t\right) \\
&&&= (1-t)^{\frac{n}{2m}-1} \left ( 4 - (4 + \frac{2n}{m})t \right) \\
\end{align*}

Therefore $\displaystyle t = \frac{2m}{2m+n}$  and $\displaystyle A = 4\cdot \frac{2m}{2m+n} \cdot (1 - \frac{2m}{2m+n})^{n/2m} = \frac{8m}{2m+n} \cdot \left ( \frac{n}{2m+n}\right)^{n/2m}$