Reading an old calculus book I have found this interesting exercise:
"Show that the rectangular box of maximum volume inscribed in a sphere of radius $r$ is a cube"
I solved this using Lagrange. But I think this can be solved using partial derivatives. However, I'm not entirely sure about this. So I would like to know if this is possible.
Best Answer
Sphere is given by $x^2 + y^2 + z^2 = r^2$.
For the rectangular box with center at the origin,
$V = 8 xyz = 8xy \sqrt{r^2 - x^2 - y^2}$
$\frac{\partial V}{\partial x} = 8y \sqrt{r^2 - x^2 - y^2} - \frac{8x^2y}{\sqrt{r^2 - x^2 - y^2}}$
To find critical points, we equate it to zero.
As $\sqrt{r^2 - x^2 - y^2} \ne 0$ in a sphere, we get $8y (r^2 - x^2 - y^2) - 8x^2y = 0$
Again as $y \ne 0$, $2x^2 + y^2 = r^2$.
Similarly from the other partial derivative (wrt $y$), we get $2y^2 + x^2 = r^2$
Solving them for $x, y$ and then plugging into equation of sphere to find $z$, we will notice they are all equal and of length $\frac{2r}{\sqrt3}$.