Say I have a constrained optimisation problem where I need to minimize $f(x,y)$ subject to the constraint that $g(x,y) = c$.
The lagrangian is defined as
$$\mathcal{L}(x,y,\lambda)=f(x,y)+\lambda (g(x,y)-c)$$
and I think we attempt to minimize it in order to find a solution to our original optimisation problem.
The intuition that I'd like to use (if it's accurate), is that the minimising lagrangian is sort of a relaxed version of the original problem, where we have a "penalty" or "subsidy" when our constraint is not satisfied (as the function might take on a smaller or larger value).
But here, aren't we permitting the equality constraint to be violated, so long as the lagragian is minimised? Why not instead use
$$M(x,y)=f(x,y) + (g(x,y)-c)^2$$
and attempt to minimise this instead? Wouldn't this make the optimal value for the second term be $g(x,y) = c$? I was thinking maybe this function would give greater "importance" to satisfying the constraint, as all other values for $(x,y)$ would lead to an increase in $M(x,y)$.
Thanks!
You have misunderstood Lagrangian multipliers.
The theorem says that if $(x_0, y_0)$ is a constrained optimum (and the gradient of $g$ is nonzero), then there is a $\lambda_0$ such that $(x_0, y_0, \lambda_0)$ is a stationary point, not necessarily a minimum, of the Lagrangian function $L(x, y, \lambda)$. The stationary condition forces $g=c$, which you can see by taking the partial derivative of $L$ with respect to $\lambda$ and setting it to zero.
In fact, the stationary point of the Lagrangian that solves the original problem is never an extremum (maximum or minimum); see this. (Even if you don’t have access, just read the first page of the preview.)
Thus your question makes little sense. In fact, all stationary points of $L$ do result in the constraint $g=c$ being satisfied.