What is the smallest volume of the ellipsoid that encloses the golden cuboid?

Question

Check a previously asked question about the golden cuboid.

http://answers.yahoo.com/question/index;_ylt=AnVg6f8cqUQzndz_n3rNSnrty6IX?qid=20070616000632AAPdr22&show=7#profile-info-sOLfGP3oaa

What is the smallest volume of an ellipsoid that encloses the golden cuboid can have? Assume that the sides of the golden cuboid are:

a = 1/Φ
b = 1
c= Φ

where Φ = (1/2)(1+√5)

Note that the volume of this golden cuboid is 1.

Answer 1

The smallest volume for an ellipsoid containing a regular parallelepiped happens when the ellipsoid circumscribes the parallelepiped (ie. all 8 vertices lie on the parallelepiped).

The axes of the ellipsoid will correspond to the lines going from the center of the parallelepiped through its faces' centers.

Thus, if the dimensions of the parallelepiped are d, e, f then the ellipsoid's equation will be something like:
(2x/d)² + (2y/e)² + (2z/f)² = 3

Hence the volume will be:
V = (4/3)π(d√3/2)(e√3/2)(f√3/2)
= (π/2)fed√3

Since we have 1/Φ, 1, and Φ

V = π√3/2