So I have a constrained optimization problem in the plane which I am able to solve. However, there is a twist to this problem. I must explain why the extrema I've found are the global maximum and minimum. The problem is along the following lines (I can't give the full problem because it is homework and I don't want it solved for me): $$ f(x, y) = .... \quad g(x,y) = ... = 5 $$ $f$ is the function I'm optimizing and $g$ is the constraint. Both are continuous and simple polynomial functions of $x$ and $y$ with small powers (that happens to be even).
My thinking is that when I plot $g(x,y) = 5$ as an implicit curve, it forms a loop and is compact. Therefore, the points I've found on it, are the global maximum and minimum. Is that correct? If so, how do I show that $g(x,y)$ is looping/compact?
It seems that the function $g:\>{\mathbb R}^2\to{\mathbb R}$ is $C^1$, hence continuous. This implies that the feasible set $S:=g^{-1}\bigl(\{5\}\bigr)$ is closed. For a proof that $S$ is compact you have to make sure that $S$ is bounded as well. Often this can be verified by inspection.
But this is not all: You have to check whether $\nabla g(x,y)\ne(0,0)$ in all points of $S$. Only then you can trust that the candidate list produced by Lagrange's method contains ${\rm argmax}$ and ${\rm argmin}$ of $f\restriction S$. If $S$ contains points $(x,y)$ with $\nabla g(x,y)=(0,0)$, say a selfintersection of the curve $S$, then you have to add these points to the candidate list.