Let's say with the constraint function $$g(x,y) = x^2 + y^2 -1 > 0, $$ I am maximizing $$f(x,y) = x + y -1 .$$ Intuitively, since the constraint function does not provide a finite constrained region, there is no solution, and $f(x,y)$ can infinitely increase.
However, if I just plainly proceed with Lagrange multiplier calculation, $$L(x,\lambda) = (x + y -1) +\lambda(x^2 + y^2 -1)$$ I get $$y,x = -\sqrt{2}/2, \lambda = 1/\sqrt{2}$$ And this still satisfies KKT conditions $$g(x) \geq 0 , \lambda \geq 0, \lambda g(x) = 0.$$
I expected that I would have some result that conflicted KKT conditions, but nothing was violated.
Does this mean that Lagrange multiplier under KKT condition can produce a solution that cannot be solved, without giving any indication of intractability?
Lagrange's method finds the conditionally stationary points of $f$ on the unit circle $S^1$. That's all. Whether these points give rise to a global min or max of $f$ on the feasible set $\bigl\{(x,y)\>\bigm|\, x^2+y^2\geq1\bigr\}$ is a totally different matter. This has to be discussed in a subsequent analysis of the overall findings.
To give an example: If the problem would have been to find the min of the function$g(x,y):=x^2-x y+2y^2$ on the same feasible set, then the points found by Lagrange's method would be useful.