I have solved the following problem and I would appreciate feedback on my solution, thanks.
A rectangular box is inscribed in a hemisphere of radius r.
Find the dimensions of the box of maximum volume.
What I have done:
Considering the box with the center of its base at the origin, we have the volume $V(x,y,z)=2x\cdot 2y\cdot z=4xyz=4xy\sqrt{r^2-x^2-y^2},\ x,y,z>0$
$$Df(x,y)=4\begin{bmatrix}y\sqrt{r^2-x^2-y^2}-\frac{x^2y}{\sqrt{r^2-x^2-y^2}} & x\sqrt{r^2-x^2-y^2}-\frac{xy^2}{\sqrt{r^2-x^2-y^2}}\end{bmatrix}=\begin{bmatrix}0 & 0\end{bmatrix}\Leftrightarrow \begin{cases}y\sqrt{r^2-x^2-y^2}\left(1-\frac{x^2}{r^2-x^2-y^2}\right)=0\\ x\sqrt{r^2-x^2-y^2}\left(1-\frac{y^2}{r^2-x^2-y^2}\right)=0\end{cases}\Leftrightarrow \begin{cases}y=0 &\text{ or } x^2+2y^2=r^2\\ x=0 &\text{ or } 2x^2+y^2=r^2\end{cases}.$$ We have to discard the solutions $x=0$ or $y=0$ since in that case we wouldn't have a box.
So it must be $x^2+2y^2=r^2$ and $2x^2+y^2=r^2$ thus $x^2+2y^2=2x^2+y^2\Leftrightarrow x^2-y^2=0\Leftrightarrow x=y\Rightarrow z=\sqrt{r^2-x^2-y^2}=\sqrt{r^2-2x^2}=\sqrt{r^2-2y^2}.$
Since $x^2+y^2=3x^2=r^2$ we have $\fbox{$x=\frac{r}{\sqrt{3}}$}$, similarly $x^2+2y^2=3y^2=r^2$ so $\fbox{$y=\frac{r}{\sqrt{3}}$}$ and finally $\fbox{$z=$}\sqrt{r^2-x^2-y^2}=\sqrt{r^2-2x^2}=\sqrt{r^2-2y^2}=\sqrt{r^2-2\left(\frac{r}{\sqrt{3}}\right)^2}=\sqrt{r^2-\frac{2r^2}{3}}=\sqrt{\frac{r^2}{3}}=\fbox{$\frac{r}{\sqrt{3}}$}.$
Thus the box of maximum volume inscribed in a hemisphere of radius $r$ is a box with a square basis with side length $2x=2y=\frac{2r}{\sqrt{3}}$ and height $z=\frac{r}{\sqrt{3}}$ hence with volume $V=2x\cdot 2y\cdot z=\frac{4r^3}{\sqrt{27}}.$