Lagrange multiplier when decisions variables are not in the same set 
Find the maximum of $2x+y$ over the constraint set $$S = \left\{ (x,y) \in \mathbb R^2 : 2x^2 + y^2 \leq 1, \; x \leq 0 \right\}$$

I want to use Lagrange multipliers to find the optimal solution. However, Lagrange requires $\vec x \in A$. In our case $R$, however $x$ can only be negative or zero. How can I get rid of this constraint? My idea is to do $x=w-z, w-z \le 0, w,z \in R$, but I am not sure if this is the right way to do it.
 A: Formulating the problem a bit better, you have the following :

Find the maximum of $f(x,y) = 2x + y$ over  $S = \{(x,y) \in \mathbb R^2 : 2x^2 + y^2 \leq 1, \; x \leq 0\}$.

Recall that one of the most important Lagrange Multiplier methods is the Kuhn-Tucker Lagrange method. The KTL method calculates the total minimum of a function. To yield the minimum from the function $f(x,y)$, simply take the function $-f(x,y)$. The point will then be the maximum for $f(x,y)$.
Thus, naming our role functions :
$$f_0(x,y) = -f(x,y) = -2x-y, \; \; f_1(x,y) = 2x^2+y^2-1, \; \; f_2(x,y) = x$$
You are now called to solve the system :
$$\begin{cases} \nabla f_0 + \lambda_1 \nabla f_1 + \nabla f_2 = 0 \\ \lambda_1 f_1 = 0\\ \lambda_2 f_2 = 0\end{cases} \Rightarrow\begin{cases} \begin{bmatrix} -2\\-1\end{bmatrix} + \lambda_1\begin{bmatrix} 4x \\2y\end{bmatrix} +\lambda_2 \begin{bmatrix} 1 \\ 0 \end{bmatrix}= \begin{bmatrix} 0 \\ 0 \end{bmatrix} \\ \lambda_1(2x^2+y^2-1) = 0 \\ \lambda_2 x =0\end{cases}$$
$$\implies$$
$$\begin{cases} -2 + 4\lambda_1x +\lambda_2 & = 0 \\ -1 + 2\lambda_1y &=0 \\\lambda_1(2x^2+y^2-1) & = 0 \\ \lambda_2x & = 0\end{cases}$$
Now, yield cases for $\lambda_1, \lambda_2 = 0 \; \text{or} \; \neq 0$ or any combination of them, taking into account possibilities for $x,y = 0 \; \text{or} \; \neq 0$ too. You will then find some possible minimum points. The point for which the value of $f_0$ will be the smaller will be your minimum points, thus  maximum for $f$.
A: You can solve this problem introducing convenient slack variables in order to eliminate inequalities. The inequalities
$$
2x^2+y^2 \le 1 \rightarrow 2x^2+y^2-1+\epsilon_1^2 = 0\\
x \le 0 \rightarrow x+\epsilon_2^2 = 0
$$
and the lagrangian thus is
$$
L(x,y,\lambda_1,\lambda_2,\epsilon_1,\epsilon_2) = 2x+y+\lambda_1(2x^2+y^2-1+\epsilon_1^2)+\lambda_2(x+\epsilon_2^2)
$$
the stationary condition $\nabla L = 0$ gives the conditions
$$
\left\{
\begin{array}{rcl}
 \lambda_2+4 \lambda_1 x+2& = & 0 \\
 2 \lambda_1 y+1& = & 0 \\
 \epsilon_1^2+2 x^2+y^2-1& = & 0 \\
 \epsilon_2^2+x& =& 0 \\
 \epsilon_1 \lambda_1& =& 0 \\
 \epsilon_2 \lambda_2& =& 0
\end{array}
\right.
$$
after solving we have
$$
\begin{array}{ccccccc}
x & y & \lambda_1 & \lambda_2 & \epsilon_1 & \epsilon_2 & 2x+y \\
 0 & -1 & \frac{1}{2} & -2 & 0 & 0 & -1 \\
 0 & 1 & -\frac{1}{2} & -2 & 0 & 0 & 1 \\
 -\frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{3}} & \frac{\sqrt{3}}{2} & 0 & 0 & -\frac{1}{\sqrt[4]{3}} & -\sqrt{3} \\
\end{array}
$$
Here when $\epsilon_i = 0$ it means that the corresponding restriction is active. 
