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I was just thinking about this studying for a biology test as for cells a greater surface area to volume ratio is desirable, so I wondered if there was a theoretically optimal shape that would maximize this ratio. I thought it might be that there is no such shape as you could just add folds to the surface indefinitely but I don't know if that is truly the case.

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    $\begingroup$ There's no upper bound on the surface area given a fixed volume (think of a very long and thin rectangular prism) and it can even be infinite (en.wikipedia.org/wiki/Gabriel%27s_horn). Surely in biology you don't just want a large surface area but also some kind of restriction on the diameter so it doesn't take too long for molecules to be transported between different parts of the cell. $\endgroup$ Commented Sep 7, 2023 at 1:32
  • $\begingroup$ Off-Topic: Assuming that you accept Darwinism, there is a separate issue. Assume (perhaps wrongly) that the greater the surface area per volume, the better, regardless of other considerations. This doesn't necessarily make it game over, regardless of the complexity in evolving from (for example) an ameoba to a human. Some chain of mutations must occur that lead to the high surface area shape. Absent the occurrence of such mutations, it becomes irrelevant how desirable such a cellular shape would be. $\endgroup$ Commented Sep 7, 2023 at 2:54
  • $\begingroup$ This was more of a thought inspired by biology rather than a question about biology itself but thank you for your answers it is very helpful. $\endgroup$ Commented Sep 7, 2023 at 12:28

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One way to show that there is no maximum surface area associated with a given volume is to consider very long cylinders (biologically, roughly the shape of a worm). The volume $V$ of a cylinder of radius $r$ and height (or length if laid horizontally) $h$ is:

$$V = \pi r^2h$$

For a given volume, say $V^*$ we can therefore express $h$ in terms of $r$:

$$h=\dfrac{V^*}{\pi r^2}$$

The surface area $A$ is:

$$A=2\pi rh+2\pi r^2$$

Substituting for $h$:

$$A= \dfrac{2V^*}{r}+2\pi r^2 > \dfrac{2V^*}{r}$$

Given the constant $2V^*$ in the numerator, we can therefore make $A$ as large as we like by choosing a sufficiently small value of $r$.

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