# Find Maximum Volume of a rectangular box that is inscribed

Find Maximum Volume of a rectangular box that is inscribed in a sphere of radius r.
The part i do not get the solutions manual got , and $2z$ as the volume. Can you explain how the volume of the cube for this problem is $v=8xyz?$
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Bertha Stark
Step 1
Recall that the volume of a box is $V=l×w×h$. The volume of the box inscribed inside a sphere where is the point in octant 1 where the box touches, the sphere will be $V=\left(2x\right)\left(2y\right)\left(2z\right)=8xyz$. Since this point lies on a sphere, it must satisfy ${x}^{2}+{y}^{2}+{z}^{2}={r}^{2}$. We have two equations in which we can get rid of one variable.
$V=8xyz$
$z=\frac{V}{8xy}$
${x}^{2}+{y}^{2}+{z}^{2}={r}^{2}$
$z=\sqrt{\left({r}^{2}-{x}^{2}-{y}^{2}\right)}$ (no $±$ because in octant 1, z is positive)
$\frac{V}{8xy}=\sqrt{\left({r}^{2}-{x}^{2}-{y}^{2}\right)}$
$V=8xy\sqrt{\left({r}^{2}-{x}^{2}-{y}^{2}\right)}$
This is just an optimization problem now.
$\frac{\partial f}{\partial x}=8\left(y\sqrt{\left({r}^{2}-{x}^{2}-{y}^{2}\right)}-\frac{\left({x}^{2}\right)y}{\sqrt{\left({r}^{2}-{x}^{2}{y}^{2}\right)}}\right)=0$
$\frac{\partial f}{\partial y}=8\left(y\sqrt{\left({r}^{2}-{x}^{2}-{y}^{2}\right)}-\frac{\left({y}^{2}\right)x}{\sqrt{\left({r}^{2}-{x}^{2}{y}^{2}\right)}}\right)=0$
The solution to these sets of equations are (there's also (0, 0) but that's obvious not a maximum point)
The maximum volume therefore is:
$V=8xyz$
$=8\left(\frac{r}{\sqrt{3}}\right)\left(\frac{r}{\sqrt{3}}\right)\left(\frac{r}{\sqrt{3}}\right)$
$=\frac{8{r}^{3}}{\left(3\sqrt{3}\right)}$