Minimum length of n-dimensional cubes to cover k p

Question

let's assume that we have k points having n coordinates.

 (a11, a12, a13, ...., a1n)

 (a21, a22, a23, ...., a2n)

 .
 .

 (ak1, ak2, ak3, ...., akn)

and we're allowed to use x number of n-dimensional cubes to cover those points

(if the points are on the cube, like they are on the surface or side or vertice of the cube, or inside the cube, then we consider the point to be covered by the cube).

If k and x are fixed, and all cubes must have the same side length, can we figure out what would be the minimum side length of squares, so that they cover all the points? Cubes can overlap, and they must be parallel to the coordinate axes.

For instance, let's n=2, and k=5, x=2, and the points are (2, 0), (0, 4), (2, 2), (3, 2), (0, 8), then the minimum side length of the cubes should be 4, and the cube with vertices (0, 0), (0, 4), (4, 0), (4, 4) and one with vertices (4, 0), (4, 4), (8, 4), (8, 0) would cover all the points

I was wondering if there was a way to do it. For n=1, it's pretty trivial, and if there are well-known algorithms for n=2, n=3 cases, maybe we could extend from them.