$$\begin{align*} &\text{minimize }& f(\b{x}) \br &\text{subject to }& \b{h}(\b{x}) = \b{0}\br && \b{g}(\b{x}) \leq \b{0} \end{align*}$$
where $f:\R^n \to \R$, $\b{h}:\R^n \to \R^m$, $m \leq n $, and $\b{g}: \R^n \to \R^p $.
Let $\b{x}^* $ satisfy $\b{h}(\b{x}^*) = \b{0} $, $\b{g}(\b{x}^*) \leq \b{0} $, and let $J(\b{x}^*) $ be the index set of active inequality constraints:
$$J(\b{x}^*) \triangleq \set{ j : g_j(\b{x}^*) = 0 } $$
Then, we say that $\b{x}^* $ is a regular point if the vectors
$$\nabla h_i(\b{x}^*), \nabla g_j(\b{x}^*), 1 \leq i \leq m, j \in J(\b{x}^*) $$
are linearly independent.
An inequality constaint $g_j(\b{x}) \leq 0$ is said to be active at $\b{x}^* $ if $g_j(\b{x}^*) = 0 $. It is inactive at $\b{x}^* $ if $g_j(\b{x}^*) \leq 0 $.
Let $f, \b{h}, \b{g} \in \mathcal{ C }^1 $. Let $\b{x}^* $ be a regular point and a local minimizer for the problem of minimizing $f $ subject to $\b{h}(\b{x}) = \b{0} $, $\b{g}(\b{x}) \leq \b{0} $. Then, there exist $\b{\lambda}^* \in \R^m $ and $\b{\mu}^* \in \R^p $ such that:
- $\b{\mu}^* \geq \b{0} $
- $Df(\b{x}^*) + \transpose{ \b{\lambda}^* } D \b{h}(\b{x}^*) + \transpose{ \b{\mu}^* } D \b{g}(\b{x}^*) = \transpose{ \b{0}}$
- $\transpose{ \b{\mu}^* } \b{g}(\b{x}^*) = 0$
We call $\b{\lambda}^* $ the Lagrange multiplier vector and call $\b{\mu}^*$ the Karush-Kuhn-Tucker (KKT) multiplier vector. We refer to their components as the Lagrange multipliers and the Karush-Kuhn-Tucker (KKT) multipliers, respectively.
Observe that $\b{\mu}_j^* $ and $g_j(\b{x}^*) \leq 0 $. Therefore, the condition
$$\transpose{ \b{\mu}^* } \b{g}(\b{x}^*) = \mu^*_1g_1(\b{x}^*) + \mu^*_2g_2(\b{x}^*) + … + \mu^*_{ p }g_{ p }(\b{x}^*) = 0$$
implies that if $g_j(\b{x}^*) < 0 $, then $\mu_j^* = 0 $; that is, for all $j \not \in J(\b{x}^*) $, we have $\mu_j^* = 0 $. In other words, the KKT multipliers $\mu_j^* $ corresponding to inactive constraints are zero.