Unusual 3D Packing Problem

Question

I made up this interesting problem playing with wire sculptures:

If I have a $10 \times 10 \times 10$ clear box and inside I can put wireframe unit cubes, what's the maximum number of unit edges (or total edge length) I can get if I must have exactly $800$ unit cubes in my box? Every unit cube is clear and you can see all $12$ edges, and if two unit cubes are connected by an edge it only counts as a single edge. The unit cubes can float, too and must be on lattice points.

My guess would be to spread out the unit cubes as much as possible to get the most unique edges. Is this equivalent to spreading out the "air pockets" as well as possible? I believe this is some kind of packing problem, but I don't have any experience with those.

Answer 1

If the unit cubes are not constrained to the lattice and can float, you can make sure that none of the edges overlap, so can have $12 \cdot 800 = 9600$ edges

If the unit cubes are constrained to the lattice, with many fewer than $800$ cubes you can make sure to have all the edges. In each direction there are $11 \times 11$ edges perpendicular to each layer, for $3 \cdot 11 \cdot 11 \cdot 10 = 3630$ total edges.

Challenge for you: what is the fewest number of cubes constrained to the lattice that give all $3630$ edges?

Answer 2

You can checkerboard (in 3D) the clear box with only 500 unit cubes. This will pretty much give you every unit edge except 60 along the edges of the clear box. With just 552 (some of the remaining edges share the same unit cube) you already have all the unit edges.