Consider a bucket in the shape of a cube 1 foot on a side and filled with water.
A smaller cube shaped container, open at the top, is pushed straight down into the bucket without rotating it. At first it displaces some water which spills out of the bucket but when this container is pushed down far enough the extra water will pour into it.
If this container is very small it will be completely filled and sink to the bottom. If it is very big it will not end up with much water in it. What dimensions of this cubic container will maximize the volume that ends up inside of it.
Let the side of the smaller cube be s.
The "very small cube" will hold s^3 cubic units of water.
But the water level in the larger cube must not go below the top of the smaller cube in order to reach this limit. That happens when s^3 is indeed the amount of water at a higher level than the top of the smaller cube. But s^3 of water has also been lost outside the larger cube.
So at that point, 2*s^3 + s = 1.
s = .6
s = (1 - s) ^ (1 / 3) / 2
PRINT s, s ^ 3
quickly finds s = .4175611742406833.
At that point the cube is completely filled with the water that is at a higher level than its top. Any larger, and there'd be less water at a higher level, and so the contents would go down. The cube of this s has volume .0728048532199146.
So each dimension of the smaller cube is the solution to 2*s^3 + s = 1, and this is approximately .4175611742406833.
Posted by Charlie
on 2010-02-03 14:21:56