Tips on resolving a Lagrange Multipliers equation system I'm having a very hard time resolving the system of equations after using the Lagrange Multipliers optimization method. For instance:
The plane $ x + y + 2z = 2 $ intersects the paraboloid $ z = x^2 + y^2 $ over an ellipse. Find the ellipse points that are nearer and farther from the origin.
I know that the Lagrange equation is going to be:
$ L(x,y,z) = \sqrt{x^2 + y^2 + z^2} - L_1(x + y + 2z - 2)-L_2(z - x^2 - y^2)$
For simplicity it can be rewritten as:
$ L(x,y,z) = x^2 + y^2 + z^2 - L_1(x + y + 2z - 2)-L_2(z - x^2 - y^2)$
And that the equation system will be:
$ \frac{\partial L}{\partial x} = 2x - L_1 + 2 L_2 x = 0 $
$ \frac{\partial L}{\partial y} = 2y - L_1 + 2 L_2 y = 0 $
$ \frac{\partial L}{\partial z} = 2z - 2L_1 - L_2 = 0 $
$ \frac{\partial L}{\partial L_1} = x + y + 2z - 2 = 0 $
$ \frac{\partial L}{\partial L_2} = z - x^2 - y^2 = 0 $
Solving this system is really giving me a hard time, so instead of asking for the answer (which it could also help to see what are your procedures) I'd like to know if there are any known tips that could lead on a solution for the system.
 A: Hint.  From my experience of students doing this kind of question, here is the no.1 tip: do not miss any potential solutions.  For example, the solution of
$$xL_1=yL_1$$
is not $x=y$, it is
$$x=y\quad\hbox{or}\quad L_1=0\ .$$
It's hard to give any other general tips for solving equations like this, which will usually be non-linear.  Usually I would try to eliminate one variable at a time.  And I would never (well almost never) pursue a line of solution if it is getting difficult, I would go back to an earlier stage and try a different idea.  If the different idea turns out even worse you can always go back to the one you abandoned.

For your problem, the first two equations give
$$2x(1+L_2)=2y(1+L_2)\ .$$
First case: $x=y$.  Then use the fifth equation to get $z$ in terms of $x$, then the fourth gives a quadratic in $x$.

Second case: $L_2=-1$.  Then the first two equations give $L_1$ and the third equation gives a value for $z$; however a bit of thought shows that this value for $z$ is impossible.  Therefore only the first case gives solutions.
See if you can fill in the details.  Good luck!
A: Hint
As commented by David, some of your partial derivatives are wrong and I suppose that you have fixed them.
In any manner, the first four derivatives make a linear system. So, just solve it the manner you want and, as a result, $x,y,z,L_1$ will simply express as function of $L_2$. Now, plug these expressions in the fifth derivative : this will lead to an equation in $b$ you can solve for $b$. Now, go backward and get the value of each parameter.
I am sure that you can take from here.
