I have a Parallelepiped which has a 3D size x, y and z I would like to know how could I get the size (x, y, z) of the non-uniform Sphere which can contains the Parallelepiped.
Here's an illustration but with a Parallelepiped and a non-uniform Sphere that fits the Parallelepiped shape.
Thanks a lot !
If you think to an ellipsoid with axes orthogonal to the faces of the parallelepiped (see figure below), here is a solution.
Consider the coordinate system with the origin at the center of the parallelepiped "box", and axes $Ox, Oy, Oz$ orthogonal to the faces of the box, piercing them at their center.
Let $2x_0, 2y_0, 2z_0$ be the resp. side lengthes of the box. Thus the coordinates of the vertices of the box are $(\pm x_0, \pm y_0, \pm z_0)$.
Then the ellipsoid defined by the following equation :
$$\dfrac{x^2}{x_0^2}+\dfrac{y^2}{y_0^2}+\dfrac{z^2}{z_0^2}=3$$
(https://en.wikipedia.org/wiki/Ellipsoid) passes through the 8 vertices and includes the box (see proof below).
The endpoints of this ellipsoid can be found by setting two of the three variables to zero, giving
$$(x_e=\pm \sqrt{3}x_0, 0, 0), \ \ \ \ (0, y_e=\pm \sqrt{3}y_0, 0), \ \ \ \ (0, 0, z_e=\pm \sqrt{3}z_0)$$
Proof of inclusion : the interior of the ellipsoid is convex. The box is the convex hull of its vertices; as such, it is included into any convex set containing these points.
Case where $x_0=3, \ y_0=2, \ z_0=1$.