Maximum volume of parallelepiped Find the dimensions of the parallelepiped of maximum volume circumscribed by a sphere of radius R.
I would normally be familiar with this using lagrange multipliers, but how do I do this?  It probably helps that I do not know the volume of a parallelepiped.  Thanks!
 A: By symmetry, you can suppose edges parallel to the coordinate axes. Let be $(x,y,z)$ the coordinates of a vertex. By hypothesis, $x^2+y^2+z^2=R^2$. The volume is $V(x,y,z)=|2x\,2y\,2z|$. Study $f=V^2$ with restriction $x^2+y^2+z^2=R^2$.
A: Hint:  The answer is obtained when all vectors spanning the parallelopiped are on the boundary of the ball of radius $R$ (because if you make a vector longer, then the parallelopiped spanned gets bigger). So you just need to maximize the volume such that each spanning vector has length $R$.  For this, you can use induction and use the fact that the volume of a parallelopiped is equal to area of the base times the height, where base area is determined by leaving out one vector and computing the volume of the resulting parallelopiped in one dimension lower, and the height is the length of the projection of the left out vector onto the orthogonal complement of the span of remaining vectors,. If you fix all but one vector, so the base area is fixed, how do you maximize the height? Use induction to prove that all vectors spanning the parallelopipied form an orthogonal basis when you maximize volume, and go from there to get the answer.
