The volume of a cylinder equals $V$ cubic inches, where $V$ is a constant. Find the proportions of the cylinder that minimize the total surface area.

Since we want to minimize the surface area, we want to find the critical points, which are $r=0$ (where $\frac{dS}{dr}$ is undefined) and $r=h/2$ (where $\frac{dS}{dr}$ is equal to zero). We can immediately ignore $r=0$ as being a possible solution because $V$ is a non-zero value. This leaves $r=h/2$, the book's answer to this problem, or in other words, the surface area is minimized when the radius is equal to half the height. But how do I know that the radius equalling half the height minimizes and not maximizes the surface area? The book's explanation for this is:

With $V$ fixed, we can choose $r$ and $h$ so as to make $S$ as large as we like.

How can volume be fixed but surface area be infinite. When I look this up I find Gabriel's Horn, but the 3D figure we are talking about here is a cylinder.

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    $\begingroup$ The area is not infinite, but you have have cylinders of arbitrarily large area. It's like saying that there are arbitrarily large natural numbers; none of them is infinite. $\endgroup$ – saulspatz Jun 30 '19 at 14:16
  • $\begingroup$ No one has said the surface area is infinite. Saying "we can make the surface area as large as we like" means it is unbounded, not infinite. $\endgroup$ – user247327 Jun 30 '19 at 14:17
  • $\begingroup$ To know whether $r=h/2$ "minimizes and not maximizes the surface area", test by first derivative or second derivative of $S$. $\endgroup$ – peterwhy Jun 30 '19 at 14:19
  • $\begingroup$ @saulspatz I get what you are saying and that's actually what I thought at first, but the whole point of the book's statement is to disprove that there is no such thing as a maximum surface area, which is why $r=h/2$ is automatically a minimized value. $\endgroup$ – user532874 Jun 30 '19 at 14:23
  • $\begingroup$ @peterwhy how can I even use those tests when no values are given $\endgroup$ – user532874 Jun 30 '19 at 14:24

Take a piece of clay and roll it into a cylinder. Say it has length $\ell$ and radius $r$. Then its volume is $\ell\cdot \pi r^2$. And its surface area (not counting the two ends) is $\ell\cdot 2\pi r$.

Now roll out your cylinder so that it becomes longer and thinner. Say it's only half as thick as before, the radius has decreased to $\frac r2$. But of course the volume must be the same because the amount of clay is the same. Since the $\pi r^2$ is one-fourth as big, the length must be four times as big to make it work out. The cylinder is half as thick, but four times as long.

That means that the new surface area, $4\ell\cdot 2\pi \frac r2$, is twice as large as it was. Which makes sense: the cylinder is longer and thinner, so more of it is at the surface.

Now roll it to be half as thick again. The volume is still the same, but again the surface area doubles. Think about how quickly a thin clay snake will dry out, compared with a thick chunky cylinder. Why? Because it has so much more surface area for the same volume of clay.

You can keep rolling the snake, and increase the surface area as much as you want. You can't maximize the surface area, because you can always double the surface area by rolling the snake to be half as thick and four times as long.

(Other examples to consider: You can take the same clay cylinder and smear it into a thin flat sheet that covers the whole earth. Its surface area will be the same as the earth's! But its volume is the same as it always was.

A kitchen sponge has a small volume, but a very big surface area. As you add more holes to the sponge, the surface area increases even though the volume of the sponge is getting smaller.)

It's important to notice that at no point does the snake have an infinite surface area. We can roll it out as thin as we like, and make the surface area as large as we like, but it's always a snake with finite length and finite surface area. What if the radius of the snake goes all the way to zero? Well, that doesn't really make sense. It's not a cylinder any more, it's a line, and a line doesn't have a surface area.

But we can use this rolling process to make a shape that does have an infinite surface area! Start with the same cylinder, and instead of rolling out the whole thing to be half as thick, roll out just the right half, leaving the left half the way it was. The left half's surface area stays the same, and the right half's surface area doubles, so the total surface area has increased from $\frac12+\frac12$ to $\frac12 + 2\cdot\frac12 = 1.5$. The surface area has increased by 50%, and in particular the right half has as much surface area as the original cylinder.

Now take just the right half (which is now just as long as the original cylinder was) and roll out its right half. Again the volume of the part we roll out doubles, so the total surface area now goes from $\frac 12 + \frac12 + \frac12 $ to $\frac12 + \frac12 + 2·\frac12 = 2$.

We can keep doing this as long as we like, making this spindle thing longer and longer, and again we can make the surface area as large as we want. But this time it's a little different than when we rolled the whole snake. There, to make the snake infinitely long, we would have had to make its radius go to zero, and then we wouldn't have a solid figure any more. But no part of the spindle has a radius of zero. It gets thinner and thinner as you go along it to the right, but every part of it has a positive, nonzero thickness. So you can take the spindle all the way to infinity, and then it has an infinite surface area, but its volume stayed the same the whole time.

This spindle is Gabriel's Horn.

  • $\begingroup$ How come when I look up "infinite surface area, fixed volume", only Gabriel's horn shows up? You've proved that other 3D shapes can also take on this characteristic. $\endgroup$ – user532874 Jun 30 '19 at 14:31
  • $\begingroup$ @user532874: as others have pointed out, none of these cylinders has infinite surface area. Please take a while to digest this fact. $\endgroup$ – TonyK Jun 30 '19 at 14:33
  • $\begingroup$ @TonyK MJD says "You can't maximize the surface area" $\endgroup$ – user532874 Jun 30 '19 at 14:41
  • $\begingroup$ @user532874: Yes, that's true. You can make the surface area 100, or 1000, or 10000, or... But you can't make it infinite. Please take a little bit longer than ten minutes to digest this fact. $\endgroup$ – TonyK Jun 30 '19 at 15:22
  • $\begingroup$ @TonyK I see what you mean now. I was not being careful with the word infinite. $\endgroup$ – user532874 Jun 30 '19 at 15:31

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