Consider a container in the shape of a unit cube. It is rigid and has a tightly sealing lid. Pour into it a certain volume of water and close the lid.

For any given orientation of the cube the water and air within will have a boundary surface.

For a given volume, V, how can this surface be minimized?

Only volumes less than or equal to 1/2 need be considered as the portion of the interior that's air can be interchanged with the portion that's water.

There are apparently 4 cases to consider: 1, 2, 3 or 4 vertices are below the water line. From what I can imagine, 1, 2 or 4 vertices are possible for any volume of water, but 3 requires a specific range.

Posted by Charlie on 2009-12-07 11:25:39