Suppose I have a solution $x^*\in\mathbb{R}^n$ to the following problem
\begin{align*} \text{minimize}_{x}& \sum_{i=1}^n f_i(x)\\ \text{subject to}\quad &g_i(x) = 0\,\,i=1,\ldots,m\\ &h_i(x) \le 0\,\,i=1,\ldots,p \end{align*}
How would I go about computing the Lagrange multiplier $\lambda_i$ associated with the equality constraint $g_i(x) = 0$?
I know that the Lagrangian is \begin{align*} \mathcal{L}(x,\lambda,\mu) = \sum_{i=1}^n f_i(x) - \sum_{i=1}^m\lambda_ig_i(x) - \sum_{i=1}^p\mu_ih_i(x) \end{align*} but I'm unsure where to go from here in order to actually get a $value$ for $\lambda_i$ given the optimal solution $x^*$.
You have an inequality as a constraint. So I would use the Kuhn-Tucker conditions (KKT).
But you have an equalility, too. It can be split as follows:
$g_i(x)=0 \Leftrightarrow \begin{cases} g_i(x) \leq 0 \\ \\ g_i(x) \geq 0 \end{cases} \Leftrightarrow \begin{cases} g_i(x) \leq 0 \\ \\- g_i(x) \leq 0 \end{cases} $
$max \ \mathcal L=-\sum_{i=1}^n f_i(x)+\sum_{i=1}^m \lambda_ig_i(x)-\sum_{i=1}^m \Theta_ig_i(x)+\sum_{i=1}^p \mu_ih_i(x)$
By using KKT, $\mathcal L$ will allways be maximized. So I take here $-f_i(x)$ to minimize $f_i(x)$.
greetings,
calculus