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?
(In reply to
some observations by Charlie)
For 1/6 cubic units or less AND 5/6 cubic units or more, the minimized surface should be as an equilateral triangle with the cube oriented such that the axis of two opposing vertices are perpendicular to the horizontal plane.
For 1/6 and 5/6 cubic units of volume, the surface area would be SQRT(3)/2 square units [approx. 0.866025].
In the given orientation, for volumes between 1/6 and 5/6 cubic units, the surface area would be hexagonal, with 1/2 volume having a surface area of a regular hexagon [3/4*sqrt(3) ~= 1.299038 square units]. This is larger than when the cube is orientated with one face flat on the horizontal surface, with the surface area of a regular square at 1 square unit. Thus at some point between 1/6 to 1/2 and 1/2 to 5/6, the 1 square unit surface area, with the cube face on the horizontal surface, may be the minimum.
Edited on December 10, 2009, 7:12 am

Posted by Dej Mar
on 20091208 21:09:49 