# What height h and base radius r will maximize the volume of the cylinder if the container in the shape of a right circular cylinder with no top has surface area 3pi ft^2?

Mar 27, 2015

The maximum volume occurs when $r = 1 \text{ ft}$ and $h = 1 \text{ ft}$.

Set-Up (find the function to optimize)
For a cylinder the volume is $V = \pi {r}^{2} h$

And for a cylinder with no top, the surface area is $A = \pi {r}^{2} + 2 \pi r h$

Given the area is $3 \pi$, we can express the volume using one variable instead of two.
$A = \pi {r}^{2} + 2 \pi r h = 3 \pi$.

Solving for $h$ looks easier than solving for $r$, so let's try it that way
(I now it will work because I've been doing this for years. But a student isn't sure.)

$\pi {r}^{2} + 2 \pi r h = 3 \pi$.
leads to $h = \frac{3 \pi - \pi {r}^{2}}{2 \pi r} = \frac{3 - {r}^{2}}{2 r}$ (domain: $r > 0$)

Substituting in the formula for volume, we get:

$V = \pi {r}^{2} \left(\frac{3 - {r}^{2}}{2 r}\right) = \frac{\pi}{2} \left(3 r - {r}^{3}\right)$ (domain: $r > 0$)

This is the function we've been asked to maximize.

Optimizing the function

$V ' = \frac{\pi}{2} \left(3 - 3 {r}^{2}\right) = \frac{3 \pi}{2} \left(1 - {r}^{2}\right)$

$V ' = 0$ at $r = \pm 1$. we note that $- 1$ is not in the domain , so the only critical point is $r = 1$

$V ' ' \left(r\right) = - 3 \pi r$, so $V ' ' \left(1\right) < 0$ and the second derivative test tells us that V(1) is a local maximum. The fact that there is only one critical point allows us to change the word "local' to 'global'.

Now we re-read the question to decide how to answer it. We were asked for $r$ and $h$ to get maximum volume.
When $r = 1$ we use the substitution above to see that $h = \frac{3 - {\left(1\right)}^{2}}{2 \left(1\right)} = \frac{2}{2} = 1$
The maximum volume occurs when $r = 1 \text{ ft}$ and $h = 1 \text{ ft}$.