I have a very simple question regarding the Lagrange multiplier. Everything can be seen in the most basic setting of two functions $F : \mathbb{R}^2 \to \mathbb{R}$ and $g : \mathbb{R}^2 \to \mathbb{R}$.
Thus, given the maximization problem
\begin{align} \max_{x, y}\ & F (x, y) \\ \text{sub}\ \ & g(x, y) = m , \end{align}
I always found the corresponding Lagrangian written as
\begin{equation*} L (x, y , \mu) = F (x, y) + \mu [m - g(x,y)] \end{equation*}
or
\begin{equation*} L (x, y , \mu) = F (x, y) - \mu [g(x,y) - m]. \end{equation*}
Thus, the question is if it possible to write the Lagrangian as
\begin{equation*} L (x, y , \mu) = F (x, y) + \mu [g(x,y) - m], \end{equation*}
the point being that when we take $L_\mu = 0$, we still have nothing more than the original constraint.
Thank you in advance for any feedback.
Yes, your way of writing the Lagrangian is also valid. The Lagrangian multiplier term can be either added or subtracted, it doesn't matter. You can also use a fourth way of writing the Lagrangian:
$$L(x,y,\mu)=F(x,y)-\mu[m-g(x,y)]$$
All that matters is that the term multiplied by $\mu$ gives $m=g(x,y)$ when equated to zero.