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I have a set $S=\{(x,y,z) \in \mathbb{R}^3 \mid g(x,y,z)=x^2+y^2-z^2 = 1\}$ on which we have a function $f:S \rightarrow R$ defined by $$ f(x,y,z)=x+y+z^2 $$ I have to find all global minima and maxima. By Lagrange, I have found the global minima to be at $(x=-\sqrt{\frac{1}{2}}, y=-\sqrt{\frac{1}{2}}, z=0)$. If $z \neq 0,$ I get a complex solution, which is not what we are looking for.

Before solving I have to argue that such a minimum/maximum exist or does not exist. I know that a continious function on a compact set assume its minima and maxima, but $S$ is unbounded in our case and therefore not compact. Talking about maxima I have shown that a maximum cannot exist by contradiction. Assuming that there exists a maximum with some value, and then choosing another point based on this value, that satisfies $g=1,$ which turns out to have a bigger value according to $f.$

How do I argue that the minima does exist and is there an easier way to show that there are no global maximum? I guess I have to show that $f$ is closed and bounded from below, subject to the condition, because then a minima will exist, but I cant figure out how. I know from the Hessian that $f$ is convex while $g$ is not convex, can I use that in any way?

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For existence of minima, you have to show that $f(x,y,z)\to\infty$ if $\|(x,y,z)\|\to\infty$. Then sets of the type $\{(x,y,z):f(x,y,z)\le C\}$ are bounded.

Let $(x,y,z)$ satisfy the constraint. Then $z^2=x^2+y^2-1$ and $$ f(x,y,z)= x+y+z^2 = x+y+ \frac12 z^2 +\frac12 x^2 + \frac 12 y^2-\frac12 =\frac12(x+1)^2 + \frac12(y+1)^2 +\frac12 z^2 - \frac32. $$ The right hand-side tends to $+\infty$ for $\|(x,y,z)\|\to\infty$.

To see that this is enough, take the feasible point $(1,0,0)$. Any minimum has to have a smaller value of $f$ than $1$. So we can restrict our search to all feasible points with $f\le 1$. Due to the above inequality, such $(x,y,z)$ satisfy $$ \frac12(x+1)^2 + \frac12(y+1)^2 +\frac12 z^2 - \frac32 \le 1 $$ or $$ (x+1)^2 + (y+1)^2 + z^2 \le 5, $$ that is, they are in a circle around $(-1,-1,0)$ with radius $\sqrt 5$. The set of such feasible points is compact, so we get existence of minimum.

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  • $\begingroup$ Is it possible to say that the rewritten function is convex due to its Hessian matrix and thefore assumes a minimum (is it general that convex functions has a minimum)? All the above makes sence, but I would like to throw in as many theorems and sentences as possible. The fact that the rewritten f(x,y,z) tends to infinity is that enough to prove that there is no global maxima? $\endgroup$
    – Mathman
    Oct 1, 2020 at 9:05
  • $\begingroup$ Convexity of $f$ is of no use as the feasible set is not convex. Also $\exp(x)$ has no minimum. and yes that property implies non-existence of maximum. (This could be achieved much easier using the feasible points $(1,n,n)$). $\endgroup$
    – daw
    Oct 1, 2020 at 9:14

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