extrema of funcion $f(x,y,z)=x+2z$ and $M=\{[x,y,z]\in\mathbb{R}^3:x^2+2y^2=4,z+y\le 1\}$. I found out that M is not bounded from below so it does not have minimum or infimum. But how do I find maximum? I tried to use lagrange multipler and got:
$\nabla f-\lambda \nabla$g=0,where $g(x,y,z)]=x^2+2y^2-4$
$$1-\lambda2x=0$$
$$0-\lambda4y=0$$
$$2-\lambda \cdot 0=0$$
I am stuck here, the third equation is obviously wrong. So how do I continue? 
 A: As coffeemath noted in the comments, you're actually given (and would have to account for) two constraints here to jump into applying Lagrange multipliers directly. An example of how to do this can be found here towards the end of the page. So, instead of just following along the usual algorithm, I thought I'd show you how a little bit of thinking about the functions in this case allows you to reduce the problem to maximizing a two variable function $f_1(x,y)$ subject to a single constraint.
Note that for a fixed $x,y$ with $x^2 + 2y^2 = 4$, if $z_1 < z_2$ then $f(x,y,z_1) < f(x,y,z_2)$. So $F(z) = f(x,y,z)$, for fixed $x$ and $y$ satisfying the constraint, is maximized when $z$ is as large as possible while remaining in $M$, i.e., for $z = 1-y$. So really you may view $f$ as a function of two variables $f_1(x,y) = x + 2(1-y) = x + 2 - 2y$, and maximizing $f$ on $M$ is the same as maximizing maximize $f_1$ on the set $M_1 = \{(x,y,z) \in \mathbb{R}^2 : x^2 + 2y^2 = 4\}$. Let $g_1(x,y) = x^2 + 2y^2 - 4$. Then $\nabla f_1(x,y) - \lambda\nabla g_1(x,y) = \langle 0,0\rangle$ together with our constraint says that
\begin{align}
    1 - 2\lambda x &= 0,\\
   -2 - 4\lambda y &= 0, \text{ and}\\
    x^2 + 2y^2 &= 4.
\end{align}
The first and second equations tell us that $x,y,\lambda \neq 0$, $x = 1/(2\lambda)$, and $y = -1/(2\lambda)$. So from the last equation
\begin{align}
    4 &= x^2 + 2y^2 = \frac{1}{4\lambda^2} + \frac{2}{4\lambda ^2}\\
\implies \lambda^2 &= \frac{3}{16} \implies \lambda \in \left\{\pm \frac{\sqrt{3}}{4}\right\}.
\end{align}
Therefore $f_1$ is maximized at some $(x,y) \in \{(\pm 2/\sqrt3, \mp 2/\sqrt3)\}$. We check $f_1(\pm 2/\sqrt3, \mp 2/\sqrt3) = \pm4/3 \pm8/3 + 2$ attains a larger value at $(2/\sqrt3,-2/\sqrt3)$ than $(-2/\sqrt3,2/\sqrt3)$, and so $f$ is maximized on $M$ at $(x,y,z) = (x,y,1-y) = (2/\sqrt3,-2/\sqrt3,1+2/\sqrt3)$.
