It is well known that a sphere has the lowest surface to volume ratio. However, a related question is: What is the shape that gives the lowest surface to volume ratio if you do not include the top in the surface. That is, what is the maximal volume of an uncovered vessel of a fixed surface area?
To be more specific with the definition of top or cover -- I regard this as a continuous manifold which allow holes. The holes do not contribute to the surface, but the volume I consider is only the one up to (up defined by the direction of gravity) the first hole. The answer thus depends on the direction of the vessel. An upside-down water glass would have an infinite ratio.
For example, a cylinder whose height is equal to its radius and a missing top basis, has a volume $\pi R^2 h = \pi R^3$ and a surface (without cover) of $\pi R^2 + 2\pi R h = 3 \pi R^2$. If we fix the volume to unity, the surface in this case is $3 \sqrt[3]{\pi}$.
In comparison, a half sphere of unity volume has a smaller surface, i.e. $\sqrt[3]{18\pi}$ .
However, it's clear this is not the best shape -- a sphere cut just a little higher the equator beats the half sphere.
So, what is the shape that gives the lowest surface to volume ratio if you do not include the top in the surface?
Thanks!