The surface to volume ratio of a sphere with diameter d is given by $\frac{\pi d^2}{\frac{1}{6}\pi d^3} = \frac{6}{d}.$
The surface to volume ratio of a cube with side length d is given by $\frac{6 d^2}{d^3} = \frac{6}{d}.$
Hence the ratio is the same in both cases.
Does that contradict the known fact, that a sphere has the lowest possible surface area to volume ratio?


A sphere has the lowest possible surface area for a thing of given volume. It has the largest possible volume for a thing of given area. But surface area : volume ratio isn't a function only of the shape but of its size, and it isn't true that e.g. any sphere has a smaller SA:V ratio than any cube. A really large sphere will do better by that metric than a really small cube; a really large cube will do better than a really small sphere.

If you want to make it size-independent, you can look at something like surface area cubed : volume squared ratio. This is now dimensionless and does depend only on the shape.

For a sphere you get $(4\pi r^2)^3 : (\frac43\pi r^3)^2$ in which the $r$s cancel out and we get $64\pi^3:\frac{16}9\pi^2=36\pi:1$. For a cube you get $(6d^2)^3:(d^3)^2=216:1$. And indeed $36\pi<216$.

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  • $\begingroup$ great answer, very well described. $\endgroup$ – ThomasL Mar 3 at 20:37

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