Suppose that our surface $M\subset \Bbb R^n$ (or manifold) is defined to be the level set of a sufficiently nice function $\Phi:\Bbb R^n\to \Bbb R^m$ ($m<n$), i.e. $M = \{\mathbf x\in\Bbb R^n: \Phi(\mathbf x) =\mathbf 0 \}$. Let $f:\Bbb R^n\to\Bbb R$ be a $C^1$ function, we wish to solve the problem $$\begin{align} \text{minimize}\quad &f(\mathbf x) \\ \text{subject to}\quad &\mathbf x\in M. \end{align}$$ It is well-known that if $\mathbf x^*\in M$ is a solution to the above problem, then there exists a vector $\mathbf \lambda^*\in\Bbb R^m$ (or a real number if $m=1$) such that $$\tag{*}\label{a} \nabla_{\mathbf x} f(\mathbf x^*) = \sum_{i=1}^m \lambda^*_i \nabla_{\mathbf x}\Phi_i(\mathbf x^*). $$ This is called the method of Lagrange multipliers.
It is also very popular to rewrite the above using the auxiliary function called the Lagrange function $\Lambda:\Bbb R^{n+m}\to \Bbb R$ defined as $$ \Lambda(\mathbf x,\mathbf \lambda) = f(\mathbf x) - \sum_{i=1}^m \mathbf \lambda_i \Phi_i(\mathbf x), $$ so that \eqref{a} and the constraint $\mathbf x^*\in M$ can be recasted as $$ \nabla_{(\mathbf x,\mathbf \lambda)} \Lambda(\mathbf x^*,\mathbf \lambda^*) = \mathbf 0, $$ or written separately as $$ \partial_{ x_j} \Lambda(\mathbf x^*,\mathbf \lambda^*) = 0, \quad \partial_{\lambda_i} \Lambda(\mathbf x^*,\mathbf \lambda^*) = 0 \quad\text{for all $j=1,\dots,n$ and $i=1,\dots,m$}. $$
I have always found the formulation $\nabla_{(\mathbf x,\mathbf \lambda)} \Lambda(\mathbf x^*,\mathbf \lambda^*) = \mathbf 0$ to be a bit unenlightening. To me it is nothing but a shorthand of writing \eqref{a} and the fact that $\mathbf x^*\in M$ together in a single line, and at times even obscures the important geometric meaning that \eqref{a} conveys (from my experience of being a TA on this subject). Hence my questions today:
-Is there anything inherently interesting about the function $\Lambda(\mathbf x,\mathbf \lambda)$ itself (beside the fact that it's an auxiliary function for the Lagrange multiplier method)?
-Is there any geometric (or otherwise) intuition that is naturally associated with $\Lambda(\mathbf x,\mathbf \lambda)$?
-Is there a better answer I can give to my students when they ask "What does $\Lambda(\mathbf x,\mathbf \lambda)$ represent?" except "It's just a shorthand for writing \eqref{a}."?
I'm hopeful that there's probably some good ways of interpreting $\Lambda(\mathbf x,\mathbf \lambda)$ that I'm not aware of yet. Personally, I've been trying to avoid using it altogether if I can because I'm not sure what to make of it. Today might be a good day to fix that.