Let n ∈N, and Q n = [0,1] n let be the n-dimensional unit cube. For a nondegenerate simplex S ⊂ R n , by σ S we denote the homothetic image of with center of homothety in the center of gravity of S and ratio of homothety σ. We apply the following numerical characteristics of a simplex. Denote by ξ(S) the minimal σ > 0 with the property Q n ⊂ σS. By α(S) we denote the minimal σ > 0 such that Q n is contained in a translate of a simplex σS. By d i (S) we mean the ith axial diameter of S, i. e., the maximum length of a segment contained in S and parallel to the ith coordinate axis. We apply the computational formulae for ξ(S), α(S), d i (S) which have been proved by the first author. In the paper we discuss the case S ⊂ Q n . Let ξ n = min{ξ(S): S ⊂ Q n }. Earlier the first author formulated the conjecture: if ξ(S) = ξ n , then α(S) = ξ(S). He proved this statement for n = 2 and the case when n + 1 is an Hadamard number, i. e., there exist an Hadamard matrix of order n + 1. The following conjecture is more strong proposition: for each n, there exist γ ≥ 1, not depending on S ⊂ Q n , such that ξ(S) − α(S) ≤ γ(ξ(S) − ξ n ). By ϰ n we denote the minimal γ with such a property. If n + 1 is an Hadamard number, then the precise value of ϰ n is 1. The existence of ϰ n for other n was unclear. In this paper with the use of computer methods we obtain an equality ϰ2 = $$\frac{{5 + 2\sqrt 5 }}{3}$$ = 3.1573... Also we prove the new estimate ξ4 ≤ $$\frac{{19 + 5\sqrt {13} }}{9}$$ = 4.1141..., which improves the earlier result ξ4 ≤ $$\frac{{13}}{3}$$ = 4.33... Our conjecture is that ξ4 is precisely $$\frac{{19 + 5\sqrt {13} }}{9}$$ . Applying this value in numerical computations we achive the value ϰ4 = $$\frac{{4 + \sqrt {13} }}{5}$$ =1.5211... Denote by θ n the minimal norm of interpolation projector onto the space of linear functions of n variables as an operator from C(Q n ) in C(Q n ). It is known that, for each n, ξ n ≤ $$\frac{{n + 1}}{2}({\theta _n} - 1) + 1$$ , and for n = 1, 2,3, 7 here we have an equality. Using computer methods we obtain the result θ 4 = $$\frac{7}{3}$$ . Hence, the minimal n such that the above inequality has a strong form is equal to 4.