Extreme value problem, maximize ratio of volume to surface area 
For a cylindrical can, how to choose the ratio of the height to the radius such that the ratio of the volume to the surface area gets maximized?

The volume ist $V = \pi r^2 h$ and the surface area is $A = 2\pi r(r+h)$ so that we get
$$
 \frac{A}{V} = 2 \frac{r+h}{rh} = 2\left( \frac{1}{h} + \frac{1}{r} \right).
$$
Now for a fixed ratio of $r/h$, for example let $r/h = 0.5$ we get different values of the ratio $A/V$, for example if


*

*$r = 1, h = 2$ it is $A/V = 2\cdot 3/2$

*$r = 2, h = 4$ it is $A/V = 2\cdot 3/4$

*$r = 4, h = 8$ it is $A/V = 2\cdot 3/8$


and so on, so even for a fixed ratio we could realize different $A/V$ values, so how to solve this exercise? I guess there is something wrong with it...
 A: Of course there is no maximum value, and $\sup \frac{A}{V} = +\infty$ (just let $h$ or $r$ tend to $0$).
In case you want to maximize some meaningful ratio with respect to $\dfrac{r}{h}$, I suggest this one: $\dfrac{V^2}{A^3}$.
Indeed, we have $$\frac{V^2}{A^3} = \frac{\pi^2r^4h^2}{8\pi^3r^3(r+h)^3}=\frac{rh^2}{8\pi(r+h)^3} = \frac{1}{8\pi} \frac{x}{(x+1)^3}$$
where $x=\frac{r}{h}>0$.
By AM-GM inequality we have $$x+1 = x + \frac{1}{2} + \frac{1}{2} \ge 3\sqrt[3]{x\cdot \frac{1}{2}\cdot \frac{1}{2}} = 3\sqrt[3]{x/4},$$
which yields $$(x+1)^3 \ge 27x/4,$$
or equivalently, 
$$\frac{x}{(x+1)^3} \le \frac{4}{27}.$$
Thus, the maximum value of $\frac{V^2}{A^3}$ is $\frac{1}{8\pi} \frac{4}{27} = \frac{1}{54\pi}$, attained when $x=\frac{1}{2}$.
A: I think your ratio is the wrong way round, "Volume to Surface" is $V/A$.
Then you get
$V/A=\dfrac{rh}{2(r+h)}=\dfrac{r}{2(r/h+1)}=\dfrac{r}{2(R+1)}$
where $R=r/h$ is the ratio.
For a fixed $r$, $V/A$ is then maximized by $R\rightarrow 0$ (and $\,h\rightarrow \infty$ consequently).
