Question

show that the maximum valueof a rectangular parallelopiped enclosed in the ellipsoid x^2/a^2+y^2/b^2+z^2/c^2=1 is 8abc/3√3​

Answer 1

Answer:

Show that the volume of the largest rectangular parallelepiped that can be inscribed in the ellipsoid

x2a2+y2b2+z2c2=1

is 8abc33–√.

I proceeded by assuming that the volume is xyz and used a Lagrange multiplier to start with

xyz+λ(x2a2+y2b2+z2c2−1)

I proceeded further to arrive at abc33√.