Find Maximum Volume of a rectangular box that is inscribed

Step 1
Recall that the volume of a box is ${\displaystyle V=l\times w\times h}$. The volume of the box inscribed inside a sphere where ${\displaystyle (x,y,z)}$ is the point in octant 1 where the box touches, the sphere will be ${\displaystyle V=\left(2x\right)\left(2y\right)\left(2z\right)=8xyz}$. Since this point lies on a sphere, it must satisfy ${\displaystyle x^2+y^2+z^2=r^2}$. We have two equations in which we can get rid of one variable.
${\displaystyle V=8xyz}$
${\displaystyle z=\frac{V}{8xy}}$
${\displaystyle x^2+y^2+z^2=r^2}$
${\displaystyle z=\sqrt{(r^2-x^2-y^2)}}$ (no ${\displaystyle \pm }$ because in octant 1, z is positive)
${\displaystyle \frac{V}{8xy}=\sqrt{(r^2-x^2-y^2)}}$
${\displaystyle V=8xy\sqrt{(r^2-x^2-y^2)}}$
This is just an optimization problem now.
${\displaystyle \frac{\partial f}{\partial x}=8(y\sqrt{(r^2-x^2-y^2)}-\frac{\left(x^2\right)y}{\sqrt{(r^2-x^2{y}^2)}})=0}$
${\displaystyle \frac{\partial f}{\partial y}=8(y\sqrt{(r^2-x^2-y^2)}-\frac{\left(y^2\right)x}{\sqrt{(r^2-x^2{y}^2)}})=0}$
The solution to these sets of equations are ${\displaystyle (\frac{r}{\sqrt{3}},\frac{r}{\sqrt{3}})}$ (there's also (0, 0) but that's obvious not a maximum point)
The maximum volume therefore is:
${\displaystyle V=8xyz}$
${\displaystyle =8\left(\frac{r}{\sqrt{3}}\right)\left(\frac{r}{\sqrt{3}}\right)\left(\frac{r}{\sqrt{3}}\right)}$
${\displaystyle =\frac{8r^3}{\left(3\sqrt{3}\right)}}$