Ken's Puzzle of the Week

Rods in a Box

  1. Place 3 length=1 rods together into a configuration which maximizes the volume of the smallest box (rectangular prism) which contains the rods.  What is the configuration and resulting volume of the box?  The rods do not necessarily have to touch the box along their length; the entire structure must simply fit within the box's volume.
  2. What are the dimensions of the smallest box which can hold 3 rods regardless of their configuration?
Extension: Solve for 4 and 5 rods.

Source: Original.
Still waiting on solutions.  I think this turned out harder than expected.

If the three rods start from a single point at right angles, I worked out a box with a base touching all three rods to be:

L=W=(sqrt(3) + 1)/2
H=1/sqrt(3)
Volume=(2+sqrt(3))/(2sqrt(3)) = 1.07735
This is of course larger than a simple cube with a volume of 1.  So the cube is likely the maximum required volume.

If the three rods start from a single point to form three edges of a tetrahedron, I found the box to be:

L=W=(sqrt(3) + 1)/(2sqrt(2))
H=sqrt(2/3)
Volume=(2+sqrt(3))/(2sqrt(6))=.7618

Now, I could be wrong on these...


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