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 x be the side length of the small cube.

The maximum amount of water that the small
cube can hold is 

       h(x) = x^3                     (1)

The maximum amount of water that can spill
into the small cube is the surface area of
the water times the distance that the cube
can be pushed down

       s(x) = (1 - x^2)*(1 - x)       (2)

On the interval [0,1], h(x) is strictly 
increasing and s(x) is strictly decreasing. 
Therefore, to find the value of x which will 
maximize the amount of water we set
h(x) = s(x) and solve for x,

       x^3 = (1 - x^2)*(1 - x)

           or

       x^2 + x - 1 = 0

Thus,

       x = (sqrt(5) - 1)/2

Posted by Bractals on 2010-02-03 14:40:17