Packing Three Cubes in D-Dimensional Space

: Denote V n ( d ) the least number that every system of n cubes with total volume 1 in d -dimensional (Euclidean) space can be packed parallelly into some rectangular parallelepiped of volume V n ( d ) . New results V 3 ( 5 ) . = 1.802803792, V 3 ( 7 ) . = 2.05909680, V 3 ( 9 ) . = 2.21897778, V 3 ( 10 ) . = 2.27220126, V 3 ( 11 ) . = 2.31533581, V 3 ( 12 ) . = 2.35315527, V 3 ( 13 ) . = 2.38661963 can be found in the paper.


Main Results
The main part of this section is the proof of V 3 (5) . = 1.802803792. We use the same method as [17,18]. At the end of the section, we offer (without proof) the values of V 3 (d) for d ∈ {7, 9, 10, 11, 12, 13}.
Outline of the proof 1.
We show, that there are only two important packing configurations. Their volumes are W 1 = x 4 (x + y + z) and W 2 = x 3 (x + y)(y + z), see Figure 1. Firstly, we need to find min{W 1 , W 2 } for each {x, y, z}. The maximum from min{W 1 , W 2 } is the final result, we denote it max min{W 1 , W 2 }; 2.
Cubes with sides x . = 0.946629932, y . = 0.690148624, and z . = 0.608279275 have W 1 = W 2 = 1.802803792. We prove that this volume is sufficient for packing any three cubes with a total volume of 1 in dimension 5; 3.
We obtain an estimation of the side size of the largest cube: 0.9445 ≤ x ≤ 0.9939; 4.
Using 1 = x 5 + y 5 + z 5 and 1 > x ≥ y ≥ z > 0, we obtain constraints x 5 + y 5 ≤ 1 and x 5 + 2y 5 ≥ 1; 5. z = (1 − x 5 − y 5 ) 1/5 . Therefore, it is sufficient to work only with x and y. M is a region of {x, y} bounded by constraints from steps 3 and 4. We obtain a curve C from W 1 = W 2 , see Figure 2. Curve C divides the region M into continuous regions C 1 and C 2 , see Figure 3; 6.
We clarify: We show that the asked maximum is on curve C; (a) We use critical points for region C 1 ; We were unable to use critical points on the whole C 2 , so we gradually numerically exclude subregions. We start with comparison of maximum of subregions and 1.8 (packing with V 3 (5) > 1.8 exists).
Proof. Consider three cubes with edge lengths x, y, z in the 5-dimensional Euclidean space, where 1 > x ≥ y ≥ z > 0 and the total volume x 5 + y 5 + z 5 = 1. We are looking for the smallest volume of a parallelepiped containing all three cubes. Therefore, from several ways of packing, we can ignore the packing that leads in any circumstances to a larger volume.
Let X, Y, Z denote the cubes (sorted from the largest). We attach cubes X and Y to each other, for example, in the direction of the fifth dimension. Parallelepiped containing cubes X and Y has volume x 4 (x + y).
If we place the cube Z to the cube X in the direction of the fifth dimension, we receive volume x 4 (x + y + z). We obtain volume x 3 (x + y)(x + z) for other four directions.
If we place the cube Z to the cube Y in the direction of the fifth dimension, we receive again x 4 (x + y + z). We obtain (after appropriate shifting of the cube Y) volume , we can ignore packings that lead to the volume x 3 (x + y)(x + z).
If we start with cubes X and Z, or Z and Y, the same results are obtained. Therefore, it is sufficient to consider only two cases of packing three cubes, see Figure 1a,b. In the first case, the volume W 1 = x 4 (x + y + z) is sufficient for packing, in the second case, the volume W 2 = x 3 (x + y)(y + z) is sufficient.
(a) Packing with volume W1 (b) Packing with volume W2 Figure 1. Two cases of packing three cubes.
We need to find max min{W 1 , W 2 } under the conditions x 5 + y 5 + z 5 = 1, and For three cubes with edge lengths x 802803792. If y + z ≤ x than we can pack the cubes, as shown in Figure 1b, and volume V 2 (5) is sufficient, V 2 (5 .
x ≥ y ≥ z and . This implies that we can consider only y into x 5 + y 5 + z 5 = 1 we find the curve C: x 5 y 5 + y 10 − y 5 + (x 2 − y 2 ) 5 = 0 ( Figure 2). The interval for x can be reduced.
We are looking for max min{W 1 , W 2 }, when W 1 = x 4 (x + y + z), W 2 = x 3 (x + y)(y + z). From the condition x 5 + y 5 + z 5 = 1 we find LetC i denote the closure of the set C i . The functions W 1 , W 2 are continuous on M and the equality W 1 = W 2 holds just in the points of the curve C.
Take the point A 1 = (0.945, 0.70) ∈ C 1 . The inequality W 1 (X) < W 2 (X) holds in every point X ∈ C 1 , because of W 1 (A 1 ) < W 2 (A 1 ). Therefore, for the asked maximum holds max Take the point A 2 = (0.965, 0.65) ∈ C 2 . The inequality W 1 (X) > W 2 (X) holds in every point X ∈ C 2 , because of W 1 (A 2 ) > W 2 (A 2 ). Therefore, for the asked maximum holds max On the compact setC 1 the function (1) has its maximum in some point B.

It holds
. The equality ∂W 1 ∂y = 0 holds if x 5 + 2y 5 − 1 = 0 but the points of the curve x 5 + 2y 5 − 1 = 0 do not belong to the regionC 1 . For every point X ∈ C 1 holds ∂W 1 ∂y < 0. Therefore, the point B must lie on the curve C. We have From (2): and (x + 2y) 5 DW2y(a, b, c, d) = (a + 2c) 5 This implies that the asked maximum cannot be achieved for x ∈ [0.9475, 0.9500], see  So function (2) on the compact setC 2 must achieve its maximum in some point of the curve C. It is the same point B as above.
If we generalize considerations from the proof, we will achieve the curve C: The graph of the curve C depends on the parity of d, see Figures 5 and 6. Considering only the values 1 > x ≥ y > 0, the shape of the curve C is similar, regardless of parity, see Figure 2.
For d ≤ 10 the asked maximum is achieved on the curve C. For dimensions 7, 9 and 10 the resultsare:   Let P is intersection the constraint curve x d + 2y d − 1 = 0 and the curve C. If d = 11, then the constrained extreme on the curve C does not meet the required assumption y ≥ z. Therefore, the asked maximum must be on the constraint curve to the left of point P or on the curve C above P, see Figure 7. The same situation occurs for d = 12 and d = 13.  Figure 7. The regions C 1 , C 2 and the curve C in 11-dimensional space.

Conclusions
The issue of packing squares is an old problem and even though there are multiple partial results, it remains unresolved. We investigated a modified problem: packing three cubes in 5-dimensional space. We also calculated results for dimensions 7, 9, 10, 11, 12, 13. Considering the previous results by [17][18][19], we can say that solution is located on the curve C for dimensions 4 . . . 10. It means, that there are two (different) packings that give (the same) the largest volume.
There seems to be only a single maximal packing for dimensions greater than 10. In this packing, two smallest cubes are the same. However, the paper confirms it only for dimensions 11, 12, 13. There is a space for several improvements in our work: Is it possible to find a V 3 (d) without long numerical calculations? Is it true that two different maximum packings exist only for dimensions less than 11?
Funding: This research was partially funded by the Slovak Grant Agency KEGA through the project No. 027ŽU-4/2020.