The area to perimeter ratio and generalizations

Asked Jun 21 '15 at 02:57

Active Jun 21 '15 at 03:34

Viewed 1,196 times

I reckon (without proof) that a circle (including the interior) is the two-dimensional figure which maximizes the area-perimeter ratio.

For example, given a fixed perimeter a circle with that perimeter has $\frac{4}{\pi}$ times the area of a square with that same perimeter. Similarly, I would assume that a sphere (or possibly a ball would be the more precise term) maximizes the volume-surface area ratio.

Thus, a natural question to ask would be that an $n$-dimensional analogue of a sphere is the "shape" in $n$-space which maximizes the ratio between whatever corresponds to "volume" and "surface area" in $n$-space.

Is this true? Proof? Counter-examples?

Edit: It appears as though even the 3-dimensional case is very non-trivial. I will then slightly alter my question: has this been proven, and if it has been proven false what are the counter examples.

edited Apr 13 '17 at 12:20

Community

asked Jun 21 '15 at 02:57

MathematicsStudent1122

12,462

Of course it is true. Humps and bumps add more surface than volume. – Ross Millikan Jun 21 '15 at 03:19
Intuitively it is undoubtedly logical but I don't think intuition is very meaningful in 58-dimensions. – MathematicsStudent1122 Jun 21 '15 at 03:28
I don't have a rigorous argument, but it will come down to the fact that if $r$ is the (small) scale of the irregularity, $r^{57} \gg r^{58}$ – Ross Millikan Jun 21 '15 at 03:33
https://en.wikipedia.org/wiki/Isoperimetric_inequality#Isoperimetric_inequality_in_higher_dimensions – Chappers Jun 21 '15 at 03:35
@Chappers Thank you! That was very helpful. – MathematicsStudent1122 Jun 21 '15 at 03:37

The area to perimeter ratio and generalizations

0 Answers0