# A right cylinder is inscribed in a sphere of radius r. How do you find the largest possible volume of such a cylinder?

Nov 9, 2015

$V = \frac{4 \sqrt{3} \pi {r}^{3}}{9}$

#### Explanation:

There are several steps to this optimization problem.

1.) Find the equation for the volume of a cylinder inscribed in a sphere.
2.) Find the derivative of the volume equation.
3.) Set the derivative equal to zero and solve to identify the critical points.
4.) Plug the critical points into the volume equation to find the maximum volume.

The best place to start is by drawing a diagram. The picture below shows the cylinder inscribed in the sphere. Given the height, $h$, we can find the radius of the cylinder in terms of $r$ using the Pythagorean Theorem. Note that $h$ refers to half of the total height of the cylinder. I chose to use $h$ instead of $\frac{h}{2}$ to simplify things later on.

To find the volume of our cylinder, we need to multiply the area of the top by the total height of the cylinder. In other words;

$V = \pi \left(\text{radius of cylinder")^2 ("height of cylinder}\right)$

$V = \pi {\left(\sqrt{{r}^{2} - {h}^{2}}\right)}^{2} \left(2 h\right)$

$V = 2 \pi h \left({r}^{2} - {h}^{2}\right)$

This is our volume function. Next we take the derivative of the volume function and set it equal to zero. If we move the $h$ inside the parenthesis, we only need to use the power rule to get the derivative.

$V = 2 \pi \left({r}^{2} h - {h}^{3}\right)$

$\frac{d}{\mathrm{dx}} V \left(h\right) = 2 \pi \left({r}^{2} - 3 {h}^{2}\right) = 0$

The $2 \pi$ divides out and we are left with;

${r}^{2} - 3 {h}^{2} = 0$

After some rearranging;

${h}^{2} = {r}^{2} / 3$

Take the square root of both sides.

$h = \frac{r}{\sqrt{3}}$

This is our optimized height. To find the optimized volume, we need to plug this into the volume function.

$V = 2 \pi h \left({r}^{2} - {h}^{2}\right) = 2 \pi \left(\frac{r}{\sqrt{3}}\right) \left({r}^{2} - {\left(\frac{r}{\sqrt{3}}\right)}^{2}\right)$

Simplify.

$V = \frac{2 \pi r}{\sqrt{3}} \left({r}^{2} - {r}^{2} / 3\right)$

$V = \frac{2 \pi r}{\sqrt{3}} \left(\frac{3 {r}^{2} - {r}^{2}}{3}\right)$

$V = \frac{2 \pi r}{\sqrt{3}} \left(\frac{2 {r}^{2}}{3}\right)$

$V = \frac{4 \pi {r}^{3}}{3 \sqrt{3}}$

$V = \frac{4 \sqrt{3} \pi {r}^{3}}{9}$

This is the optimized volume for the cylinder. Its a good check to notice that $V$ is in terms of ${r}^{3}$ since volume should have cubic units. In other words, if our radius was given in term of meters, our volume units would be ${\text{m}}^{3}$