Lagrange Multiplier question (find max and min on ellipsoid) 
Find the Max and Min of $f(x,y,z)=y^2+4z^2-4yz-2xz-2xy$ under the
  constraint $g(x,y,z)=2x^2+3y^2+6z^2-1=0$

My work:
I used Lagrange theorem to get the following equation system:
$-2z-2y=\lambda 4x$
$2y-4z-x=\lambda 6y$
$8z-4y-2x=\lambda 12z$
By dividing by 2 and multiplying the first equation by $x$, the second by $y$ and the third by $z$, and then summing them all up we get that
$\lambda(2x^2+3y^2+6z^2)=\lambda=y^2+4z^2-4yz-2xz-2xy=f(x,y,z)$
I'm not sure how to carry on from here, can anyone help?
 A: So, when solving a problem through Lagrange's undetermined multiplier, you have 2 pieces that end up being considered. Your first 3 equations determine the relationship between the undetermined multiplier and your variables (x,y,z), your last is the constraint g(x,y,z)=0.
The first 3 equations you have are the derivatives and allow you to find critical values in terms of $\lambda$:
$$solution= (x_{critical}(\lambda),y_{critical}(\lambda),z_{critical}(\lambda))$$
Note, you can easily have more than 1 solution. 
With a given constraint function $g(x,y,z)=0$, we want to find the constraint evaluated at the critical values $g(x_{critical}(\lambda)_{critical},y(\lambda),z_{critical}(\lambda))=0\rightarrow g(\lambda)=0$.
Solving for the roots here allows you to plug $\lambda$ back into the critical values to calculate the extrema.
If you wanted to solve your first 3 equations I would save some time and use matrices:
$$4\lambda x+-2y-2z=0$$
$$x+(6\lambda -2)y+4z=0$$
$$-2x-4y+(8-2\lambda)z=0$$
Which can be written as:
$$\begin{bmatrix}4\lambda & -2 &-2 \\1&(6\lambda-2) &4\\-2 &-4&(8-2\lambda)\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0 \\0\\0\end{bmatrix}$$
In more general terms, our system of equation is represented as:
$$\textbf{M} \vec{v}=\vec{a}$$
$$\vec{v}_{critical}=\textbf{M}^{-1}\vec{a}$$
Leaving us with:
$$\begin{bmatrix}x_{critical}\\y_{critical}\\z{critical}\end{bmatrix}=\frac{1}{det\textbf{M}}{\begin{bmatrix}4\lambda & 1 &-2 \\-2&(6\lambda-2) &-4\\-2 &+4&(8-2\lambda)\end{bmatrix}\begin{bmatrix}0 \\0\\0\end{bmatrix}}$$
I'm going to stop it here, partly so that you have fun math to finish up, partly because my break is over.
I hope this helps.
