Drop in the bucket.

Geometry Level 3

Consider a bucket in the shape of a unit cube, open at the top and filled with water.

A smaller cube-shaped container, also open only 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 as 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, but without much water actually inside of it. If it is very big, it will not end up with much water in it, either.

What is the maximum volume of water in cubic units that the optimum container can end up with?

Note: Ignore ripples.

The answer is 0.31767219613.

1 solution

Jeremy Galvagni
Apr 19, 2018

$x$ =side length of the container. Clearly, a volume of $x^{3}$ will be immediately lost to spillage. This leaves $1-x^{3}$ anywhere inside the bucket and only a portion this will enter the cube.

Without assuming the container is filled, there will still be some volume outside of the container but inside the bucket: $x - x^{3}$

This leaves $(1-x^{3})-(x-x^{3}) = 1-x$ for the volume inside the container.

Although it seems at first that we would want to make $x$ be small, the container's volume is also given by $x^{3}$ so we also need $x$ to be big. Need to balance container size with spillage.

The solution is to set them equal (it turns out we do fill the container): $1 - x = x^{3}$

$x^{3} + x - 1 = 0$

The real root is about x = .6823278038

And so the volume of the container is $x^{3}=.6823278038^{3}=\boxed{0.31767219613}$

This was a neat problem

Steven Chase - 3 years, 1 month ago

Drop *in* the bucket.

The answer is 0.31767219613.

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Drop in the bucket.