Show that the volume of the largest rectangular parallelepiped that can be inscribed in the ellipsoid $$\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$$ is $\dfrac{8abc}{3\sqrt3}$.
I proceeded by assuming that the volume is $xyz$ and used a Lagrange multiplier to start with $$xyz+\lambda \left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}-1\right)$$ I proceeded further to arrive at $\frac{abc}{3\sqrt3}$. Somehow I seemed to be have missed $8$. Can someone please tell me where I did go wrong?
Best Answer
Maybe you need to understand the following :
Let $P=(x,y,z)$ be a point on the ellipsoid with $x,y,z\gt 0$. Take the eight different points with $$P_i (\pm x,\pm y,\pm z)$$ These points are the vertices of a parallelepiped with the side length $2x , 2y$ and $2z$. Then, the volume parallelepiped is: $$V = 2x\cdot 2y\cdot 2z = 8\cdot x\cdot y\cdot z.$$