Dimensions of a box of maximum volume inside an ellipsoid 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? 
 A: Here is a solution without calculus.
As shown in an answer to a similar question some inequalities between means can be useful here. For example, inequality between geometric mean and arithmetic mean or inequality between geometric mean and quadratic mean (a.k.a. root mean square).
We know that for any real numbers $x_{1,2,3}\ge0$ we have 
$$x_1x_2x_3 \le \left(\frac{x_1+x_2+x_3}3\right)^3. \tag{1}$$
For $x_1=x^2/a^2$, $x_2=y^2/b^2$ and $x_3=z^3/c^3$ we get
$$\frac{x^2y^2z^2}{a^2b^2c^2} \le \frac1{27}.$$
Since the volume is $V=8xyz$, we have
$$V^2 \le \frac{8^2a^2b^2c^2}{27},$$
i.e.
$$V \le \frac{8abc}{3\sqrt3}.$$
Moreover, equality in (1) is attained only for $x_1=x_2=x_3=\frac{x_1+x_2+x_3}3$ which, in our case, gives $\frac xa = \frac yb = \frac zc = \frac1{\sqrt3}$. (We get this from $x_1=x_2=x_3=\frac{x^2}{a^2}=\frac{y^2}{b^2}=\frac{z^2}{c^2}=\frac13$.)
For more about inequalities of various mean see, for example:


*

*Root-Mean Square-Arithmetic Mean-Geometric Mean-Harmonic mean Inequality at  AoPS

*Inequality of arithmetic and geometric means at Wikipedia

*Proofs of AM-GM inequality
A: 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).
