This is the question:
Show that the function $f(x,y) = x^4 + y^4$ takes maximum and minimum values along the curve $x^4+y^4-3xy = 2$.
A solution from my teacher:
Since $f$ is continuous, we only have to show that the set of solutions to the constraint is a compact set. This can be done by examining the limit $\lim_{x^2+y^2\to\infty} g(x,y)$ where $g(x,y) = x^4+y^4-3xy - 2$. Calculations shows
$$ \lim_{x^2+y^2\to\infty} g(x,y) = +\infty$$
which I guess means that there has to be a set of solutions with an upper bound (since we want $g(x,y) = 0$). However, this is where I get lost. I feel like it isn't enough.
I mean there could be no solutions (but it's not the case for this) but my main question is: how do you know the set is closed? What does a "closed set of solutions" even mean?
Is it correct to assume that it exists (we can choose) a compact set $D$ of solutions such as that every point on $g$ outside $D$ and on $\partial D$ implies that $|g(x,y| > 0$?
You do raise some valid points. First of all, note that as $g$ is continuous and $\{0\}$ is a closed set, it follows that $g^{-1}(\{0\})$ is closed. Furthermore, it is nonempty because e.g. the point $(2^{1/4},0)$ lies in $g^{-1}(\{0\})$. Now I assume you have shown that $$\lim_{x^2+y^2\to\infty} g(x,y) = +\infty.$$ By definition of the limit above, this means that there is an $R > 0$ such that $g(x,y) > 1$ whenever $x^2+y^2 > R$. In particular, any point in $(x,y) \in g^{-1}({0})$ must therefore satisfy $x^2+y^2 \leq R$ and therefore $g^{-1}(\{0\})$ is bounded. We conclude that $g^{-1}(\{0\})$ is a closed and bounded subset of $\mathbb{R}^2$ and therefore compact. The claim then follows by continuity of $f$.