Dimensions of a box of maximum volume inside an ellipsoid

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Finding the dimensions of the maximum volume box inside the ellipsoid.

I assume that the volume of a box, $V(x,y,z) = xyz$ (they did not give this to me, but this is the volume of a box right?)

Ellipsoid:

$$\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$$

and I use Lagrange multipliers to find an incorrect answer, I end up getting
$$x = \frac{\sqrt{a}}{\sqrt{3}}$$ $$y = \frac{\sqrt{b}}{\sqrt{3}}$$ $$z = \frac{\sqrt{c}}{\sqrt{3}}$$

the hint they give me is that

$$\text{Max volume} = \frac{8abc}{3\sqrt{3}}$$

Could someone tell me where I am doing this wrong?

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Probably the ellipsoid is $$ \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1 $$ and your solution becomes $x=a/\sqrt{3}$, $y=b/\sqrt{3}$, $z=c/\sqrt{3}$ which gives the correct volume (remember to multiply by $8$, because $x$, $y$ and $z$ are half the sides of the box).