Ken's Puzzle of the Week
Rods in a Box
- 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.
- 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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