# Three Cube Packing for All Dimensions
## 2 Results  
**Theorem 1**. :
1. \(V_{3}(d)=\frac{r^{1-d}}{d}\) _if_ \(d\geq 11\)_, where_ \(r\) _is the only solution of the equation_ \(2(d-1)k^{d}+dk^{d-1}=1\) _on_ \(\left(\frac{\sqrt{2}}{2},1\right)\)_._
2. \(V_{3}(d)=\frac{\frac{1}{2}+1}{r^{d}+\left(\frac{1}{r}-r\right)^{d}+1}\) _if_ \(2\leq d\leq 10\)_, where_ \(r\) _is the only solution of the equation_ \((k+1)^{d}(1-k)^{d-1}\big{(}dk^{2}+d+k-1\big{)}=k^{d}\Big{(}dk^{d+1}+dk^{d}+k^{ d}+1\Big{)}\) _on_ \(\left(\frac{\sqrt{2}}{2},1\right)\)_._ _The maximum volume is achieved by_ \(d\)_-cubes with edges_ \(x\)_,_ \(y\)_,_ \(z\)_, such that:_ 1. \(x=\left(2r^{d}+1\right)^{-1/d}\)_,_ \(y=z=rx\) _if_ \(d\geq 11\)_._ 2. \(x=\left(r^{d}+(\frac{1}{r}-r)^{d}+1\right)^{-1/d}\)_,_ \(y=rx\)_,_ \(z=(\frac{1}{r}-r)x\) _if_ \(2\leq d\leq 10\)_._  
Proof. : Let \(x\), \(y\), \(z\) be the edge lengths of \(d\)-cubes in the \(d\)-dimensional Euclidean space (\(d\geq 2\)), where \(1>x\geq y\geq z>0\) and the total volume \(x^{d}+y^{d}+z^{d}=1\). Let \(k\) and \(m\) be real numbers, such that \(y=kx\) and \(z=mx\). In the proof, we use three constraints:  
1. \(x^{d}+(kx)^{d}+(mx)^{d}=1\), i.e., \(x^{d}(k^{d}+m^{d}+1)=1\), where \(d\geq 2\), \(d\in\mathbb{N}\). 2. \(1\geq k\geq m>0\). 3. \(kx+mx>x\), i.e., \(k+m>1\). If \(k+m\leq 1\), then we pack according to Figure 1b, and the smallest \(d\)-cube does not contribute to the total volume. Three \(d\)-cubes can be packed only in two meaningful ways; see Figure 1.  
Figure 1: Two cases of packing three \(d\)-cubes. (**a**) Packing used by \(W_{1}\). (**b**) Packing used by \(W_{2}\). Let \(W_{1}=x^{d-1}(x+y+z)=x^{d}(k+m+1)=\frac{k+m+1}{k^{d}+m^{d}+1}\) be the function of the volume of the packing, as shown in Figure 1a. 2. Let \(W_{2}=x^{d-2}(x+y)(y+z)=x^{d}(k+1)(k+m)=\frac{(k+1)(k+m)}{k^{d}+m^{d}+1}\) be the function of the volume of the packing, as shown in Figure 1b. Let \(G(k,m)=min(W_{1}(k,m),W_{2}(k,m))\). The domain of the function \(G\) is bounded by the \(1\geq k\geq m>0\) and \(k+m>1\). It is shown as the triangle \(ABC\) in Figure 2. The domain is the same for each \(d\geq 2\). Our goal is to find the global maximum of \(G\) for each dimension (\(d\geq 2\)) and the edge lengths \(x\), \(y\), \(z\) for which it occurs. If \(W_{1}=W_{2}\), then \(\frac{k+m+1}{k^{d}+m^{d}+1}=\frac{(k+1)(k+m)}{k^{d}+m^{d}+1}\) gives the curve \(PB\): \(1=k(k+m)\). \(PB\) is continuous and divides the triangle \(ABC\), into two regions (see Figure 2):  
1. Region \(C_{1}\) where \(1\leq k(k+m)\) holds. Therefore, \(W_{1}\leq W_{2}\) and, consequently, \(G(k,m)=W_{1}(k,m)\). \(W_{1}\) is continuous on the region \(C_{1}\). 2. Region \(C_{2}\) where \(1\geq k(k+m)\) holds. Therefore, \(W_{1}\geq W_{2}\) and, consequently, \(G(k,m)=W_{2}(k,m)\). \(W_{2}\) is continuous on the region \(C_{2}\). The curve \(PB\) belongs to both regions. The point \(P=(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})\) is used in the proofs several times, \(G(P)=\frac{\sqrt{2}+1}{2^{1-\frac{d}{2}}+1}\). For the sake of clarity, the rest of the proof is divided into nine claims. Claim 1: The global maximum of \(W_{1}\) must occur on the boundary of the region \(C_{1}\). Proof: \(\frac{\partial W_{1}}{\partial k}=\frac{\left(k^{d}+m^{d}+1\right)-dk^{d-1}(k+ m+1)}{\left(k^{d}+m^{d}+1\right)^{2}}\), \(\frac{\partial W_{1}}{\partial m}=\frac{\left(k^{d}+m^{d}+1\right)-dn^{d-1}( k+m+1)}{\left(k^{d}+m^{d}+1\right)^{2}}\). The derivatives are never undefined. The equation \(\frac{\partial W_{1}}{\partial k}=0=\frac{\partial W_{1}}{\partial m}\) gives \(k=m\) (line segment \(PC\)). So, \(W_{1}\) has no local extremum inside of the region \(C_{1}\). The global maximum of \(W_{1}\) must occur on the boundary of the region \(C_{1}\). Claim 2: The global maximum of \(W_{2}\) must occur on the boundary of the region \(C_{2}\). Proof: \(\frac{\partial W_{2}}{\partial m}=\frac{(k+1)(k^{d}+m^{d}+1)-dm^{d-1}(k+1)(k+ m)}{(k^{d}+m^{d}+1)^{2}}\) is never undefined.