Volume of the largest rectangular parallelepiped inscribed in an ellipsoid 
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?
 A: 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: 
$$V = 2x\cdot 2y\cdot 2z = 8\cdot x\cdot y\cdot z.$$
A: Let $$h(x) = \text{maximum value}
              =8xyz +f(x^2/a^2 +y^2/b^2 +z^2/C^2+1)\tag{1}$$
where f =langragian multiplier.
Then differintiate partialy with respect to x 
$$8yz+2fx/a^2 =0\tag{2}$$
$$1/a^2 =(-8yz/2fx) $$
similarily $$1/b^2 =(-8zx/2fy)$$
$$                   1/c^2=(-8xy/2fz)$$
$$x^2/a^2 +y^2/b^2+z^2/c^2=1$$
$$\Rightarrow f=-12xyz$$
put f value in eqn(2)
$$x=a/\sqrt 3$$
similarily $$y=b/\sqrt 3$$
$$                  z=c/\sqrt 3$$
$\Rightarrow$ the largest volume of parallelopiped inscribed in ellipsoid $$=8xyz=8(a/\sqrt 3)(b/\sqrt 3)(c/\sqrt3)$$
A: You should take the dimensions of the rectangle as $2x,2y$ and $2z$ instead of $x,y$ and $z$.
