In the book Calculus by Robert A. Adams the Lagrangian function is as follows:
$$ \mathcal{L}(x, y, \lambda)=f(x, y)+\lambda{g(x, y)} $$
But in other books the function is presented as: $$ \mathcal{L}(x, y, \lambda)=f(x, y)-\lambda{g(x, y)} $$
Does somebody know why the sign is different and if it affects the methods?
Like a curiosity, I saw the positive sign in books related with physics and engineering and the negative sign in books for economics.
Thanks.
On
Initial choice of sign of $\lambda$ does not matter.
For the simplest example of finding maximum rectangle area for given perimeter length the Lagrangian
$$ 2 ( L+B) \pm ~\lambda ~LB $$
When Euler-Lagrange is used
$$\frac{2}{2}=\frac{\pm \lambda B}{\pm \lambda L} \to L=B $$
cancels $\lambda$ out along with its sign.
$\lambda$ can be replaced by any function of $\lambda$ and adjusted to suit geometrical interpretation.
In the case of an equality constraint, it is not important, though one might prefer to write $\mathcal{L}$ and $g$ in such a way that the Lagrange multiplier has a particular meaning. For example, in the utility maximization problem in economics, one may wish to write things so that the Lagrange multiplier corresponds to the marginal utility of income (rather than the negative of the marginal utility of income).
On the other hand, in the case of a maximization problem with a non-negativity constraint written in the form $g(x,y)\leq 0$, writing the Lagrangian in the form $$f(x,y)-\lambda g(x,y)$$ ensures that one of the necessary conditions for an optimum is $\lambda\geq 0$. If the Lagrangian were instead written as
$$f(x,y)+\lambda g(x,y)$$
or $g$ defined so that the constraint is $g(x,y)\geq 0$, then a necessary condition for an optimum would be $\lambda\leq 0$.
So, the necessary condition on the Lagrange multiplier(s) for inequality constraints depends on how you define the inequality constraint function(s) and the Lagrangian.