SOLUTION: Show that the rectangular solid of maximum volume that can be inscribed in a sphere is a cube?

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Question 389078: Show that the rectangular solid of maximum volume that can be inscribed in a sphere is a cube?
Answer by richard1234(7193) About Me  (Show Source):
You can put this solution on YOUR website!
Without loss of generality, let the diameter of the sphere be sqrt%283%29 (you'll see why I picked such a strange number), and that the dimensions of the rectangular solid are x, y, z. Given that:

x%5E2+%2B+y%5E2+%2B+z%5E2+=+%28sqrt%283%29%29%5E2+=+3

we want to show that for all positive x, y, z that

xyz+%3C=+1 (where 1 is the equality case and the volume of an inscribed cube).

Using the AM-GM inequality,

%28x%5E2+%2B+y%5E2+%2B+z%5E2%29%2F3+%3E=+root%283%2C+x%5E2y%5E2z%5E2%29

Since x%5E2+%2B+y%5E2+%2B+z%5E2+=+3, this becomes

1+%3E=+root%283%2C+x%5E2y%5E2z%5E2%29

Cubing both sides,

1+%3E=+x%5E2y%5E2z%5E2

1+%3E=+xyz

xyz+%3C=+1, as desired. Note that the equality case is where x = y = z = 1, which is also the equality case of AM-GM.

There is probably an optimization solution involving derivatives, however, it would involve several variables and would be fairly tedious. I'm sure many other solutions exist. This solution is the first one that came to my mind.