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`?

The maximum volume occurs when `r=1 " ft"` and `h=1 " 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 rh`

Given the area is `3 pi`, we can express the volume using one variable instead of two.
`A= pi r^2 + 2 pi rh = 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 rh = 3 pi`.
leads to `h=(3 pi - pi r^2)/(2 pi r)=(3-r^2)/(2r)` (domain: `r>0`)

Substituting in the formula for volume, we get:

`V= pi r^2 ((3-r^2)/(2r))= pi/2(3r-r^3)` (domain: `r>0`)

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

Optimizing the function

`V'= pi/2(3-3r^2)= (3 pi)/2(1-r^2)`

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

`V''(r)=-3 pi r`, so `V''(1) < 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'.

Answering the question

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=(3-(1)^2)/(2(1))=2/2=1`

Answer:
The maximum volume occurs when `r=1 " ft"` and `h=1 " ft"`.