# How to determine height of the cylinder with maximum volume engraved in a sphere with radius R?

Apr 26, 2017

h=sqrt((4R^2)/3 where $R$ is the radius of the sphere

#### Explanation:

This problem is really interesting, but will definitely require some visualization to figure out. Let's start by using the formula for the volume of a cylinder:

We know that the area of a circle is:

$A = \pi {r}^{2}$

Therefore, the volume is:

$V = \pi {r}^{2} h$

This cylinder, however, is engraved in a sphere. Its cross-sectional area and height are, therefore, restricted by the sphere as shown below:

Now, imagine cutting the sphere in half two times. After the first and second cuts, we will see:

If we focus on the cross section in the last cut, we can use the Pythagorean theorem to find a relationship between $\frac{1}{2} h$, $r$, and $R$. Specifically,

${\left(\frac{1}{2} h\right)}^{2} + {r}^{2} = {R}^{2}$
${h}^{2} / 4 + {r}^{2} = {R}^{2}$

Solving for $r$, we get:
${r}^{2} = {R}^{2} - {h}^{2} / 4$

Now that we have ${r}^{2}$ as a function of $h$, plug it back into the original volume equation:

$V = \pi \left({R}^{2} - {h}^{2} / 4\right) \cdot h$

Simplify the expression:

$V = \pi {R}^{2} h - \pi {h}^{3} / 4$

Now that we have the volume of the cylinder expressed as a function of its height, $h$, take the derivative of the volume function with respect to $h$ and set it equal to 0.

(Recall: Local maximum of a function is located where its derivative equals 0).

$\frac{\mathrm{dV}}{\mathrm{dh}} = \pi {R}^{2} - 3 {h}^{2} / 4$
$0 = \pi {R}^{2} - \frac{3 {h}^{2}}{4}$
$- \pi {R}^{2} = - \frac{3 {h}^{2}}{4}$
$- \frac{4}{3 \pi} \cdot \pi {R}^{2} = - \frac{3 \pi {h}^{2}}{4} \cdot - \frac{4}{3 \pi}$
$\frac{4 {R}^{2}}{3} = {h}^{2}$

Therefore, the value of $h$ that maximizes the volume of a cylinder engraved in a cylinder is:

$h = \sqrt{\frac{4 {R}^{2}}{3}}$

Apr 26, 2017

$h = \frac{2 \sqrt{3}}{3} R$

#### Explanation:

Assuming a cylinder with the vertical axis coincident with the $z$ axis engraved in a sphere centered at the origin and defining

$r = R \sin \theta$ Cylinder base radius
$h = 2 R \cos \theta$ Cylinder height

we have

${V}_{c} = \pi {r}^{2} h$ Cylinder volume

but

${V}_{c} = \pi {\left(R \sin \theta\right)}^{2} \left(2 R \cos \theta\right) = 2 \pi {R}^{3} {\sin}^{2} \theta \cos \theta$

so

${\max}_{\theta} {V}_{c}$ is at ${\theta}_{0}$ such that

$\frac{d {V}_{c}}{d \theta} = 2 \pi {R}^{3} \left(2 \sin \theta {\cos}^{2} \theta - {\sin}^{3} \theta\right) = 0$

or

$\left\{\begin{matrix}\sin \theta = 0 \\ 3 {\cos}^{2} \theta - 1 = 0\end{matrix}\right.$

The solutions are

$\theta = 0$ or $\theta = \pm \arccos \left(\frac{\sqrt{3}}{3}\right)$

so

$h = \frac{2 \sqrt{3}}{3} R$

NOTE: This is a maximum point because

$\frac{{d}^{2} {V}_{c}}{d {\theta}^{2}} = - \frac{8 \pi {R}^{3}}{\sqrt{3}} < 0$