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...

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Does the problem actually ask you to maximize volume for a fixed surface area, or minimize surface area for a fixed volume? Or does the problem really ask you to maximize the volume to surface area ratio? –  JimmyK4542 Jun 24 '14 at 22:59
Yes. It definitely asks to maximize the volume to surface area ratio! –  Stefan Jun 24 '14 at 23:02
In case you want to maximize some meaningful ratio then... refer to my answer ;) –  Khue Jun 24 '14 at 23:19
@Stefan: If there is no other constraint (i.e. a fixed surface area, a fixed volume, a fixed radius, a fixed base, etc.) then there is no maximum to the ratio of the volume to the surface area. That ratio is $$\frac VA=\frac{rh}{2(r+h)}$$ which can be made as large as we want. For example, if $r=h$, then the ratio is $V/A=r/4$. Thus, we can make the ratio $V/A$ as big as we wish by making $r$ big. –  robjohn Jun 24 '14 at 23:55

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).

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I am quite satisfied by your answer. But the exercise asks "for what ratio of $h$ to $r$ gets $V/A$ maximized", and what ratio should $r \to \infty, h \to \infty$ be? It could also realize every possible ratio... and of course I wrote about $V/A$ and gave the formala for $A/V$, but maximizing $V/A$ is equivalent to minimizing the reciprocal $A/V$, so thats what I head in mind. –  Stefan Jun 24 '14 at 23:22
You can see the ratio $r/h$ in this function, and it is minimized for $r/h=0$ and $r\rightarrow \infty$. I agree that this is nonsense actually. –  emcor Jun 24 '14 at 23:28
Just set $r/h$ to any constant value (e.g. $r/h=1$), and then let $r\to +\infty$ then $V/A \to +\infty$. Thus, the final answer is: $r/h$ can be arbitrary ;) –  Khue Jun 24 '14 at 23:31
r/h is not arbitrary, because the smaller r/h, the larger is V/A. So you may increase r by some amount, and increase h by even more, and get a higher V/A for this r/h ratio. –  emcor Jun 24 '14 at 23:33
@Stefan I hope the edited answer is now sufficient. –  emcor Jun 24 '14 at 23:42

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 $x=\frac{1}{2}$.

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