TITLE: How can I find the two critical points of this system of equations?
QUESTION [5 upvotes]: I'm currently trying to use Lagrange Multipliers to find the 2 critical points of the function
$$
f(x,y,z) = \frac{1}{2}x^{2}+yz+\frac{1}{3} y^{3} - z^{2}
$$
subject to 
$$
h(x,y,z) = x+y+z-2 = 0
$$
I have formed the Lagrangian for this, which is given by
$$
L= \frac{1}{2}x^{2}+yz+\frac{1}{3} y^{3} - z^{2} - \lambda (x+y+z-2)
$$
and I have worked out that
$$
L_{x}= x-\lambda = 0 \\
L_{y}= z+y^{2} - \lambda = 0 \\
L_{z}= y-2z-\lambda = 0 \\
L_{\lambda}=-x-y-z+2 =0
$$
How would I find the values of $x,y,z, \lambda$ that constitute the critical points of this system?
EDIT: I have deduced that 
$$
x=\lambda \\
y=\frac{4-\lambda}{3} \\
z=\frac{2-2\lambda}{3}
$$
Thus, since $x+y+z=2$, we must solve the equation
$$
\lambda + \frac{4-\lambda}{3} + \frac{2-2\lambda}{3} = 2
$$
for $\lambda$.
However, any $\lambda \in \mathbb{R}$ is a solution.
What am I doing wrong?

REPLY [2 votes]: You got $x$, $y$, and $z$ from below three equations
$$\begin{align}
L_{x}&= x-\lambda = 0 \\
L_{z}&= y-2z-\lambda = 0 \\
L_{\lambda}&=-x-y-z+2 =0
\end{align}$$
and put those back to one of those ($L_{\lambda}$), which of course always holds.
You should have put those back to this one
$$L_{y}= z+y^{2} - \lambda = 0$$