Let $M:=\{(x,y,z)\in \mathbb R^3: x^2+y^2+z^2\geq36\}$ and $f: \mathbb R^{3} \to \mathbb R, f(x,y,z)=\frac{x^2}{2}+\frac{y^2}{4}+\frac{z^2}{6}$
Find the Extrema and their type.
Ideas: Firstly define $h: \mathbb R^{3} \to \mathbb R$ as $h(x,y,z)=x^2+y^2+z^2-36$
We know that $M$ is not sequentially compact, using a sequence $(k+6,0,0)_{k\in \mathbb N}$ and therefore it is not compact.
Until now, I have only done constrained extrema on compact sets where we could simply find the critical points and then (based on the fact that $M$ is compact) we would know that there exists a global maximum and minimum, and then simply evaluate.
Nonetheless, in this case we know that $\lim_{(x,y,z)\to\infty}f(x,y,z)=+\infty$ and thus $f|_{M}$ has a lower bound. How can we know, however, that $f|_{M}$ "takes on" a minimum.
Question: Is there a generalized way to go about finding Extrema on non-compact sets? Do I still need to find the critical points and evaluate them or what should I do?
In this case, $\lim_{\|(x,y,z)\|\to\infty}f(x,y,z)=+\infty$ and therefore the function $f$ cannot have a global maximum. Besides, $f$ has no critical points in $M$. Therefore, $f$ has a minimum, which must be located at the boundary of $M$.