How to prove there is no minimum in this scenario?

Question

Consider the function $2x + y + z$ subject to the constraints $x^2 + y + z = 4$ and $x + 2y - z = 6$.

I run the calculations to find the max/min point and I got $P = \left(1, \frac{8}{3}, \frac{1}{3}\right)$.

Now, how can I be sure that this is a max and not a min? The two constraints do not form a compact set. How can I prove there is no global minimum here?

Answer 1

Substitute $z = 6 - x - 2y$: your objective becomes $x-y+6$ and your first constraint $x^2-x-y+2=0$. Now substitute $y = x+6 - v$ and your constraint is $x^2 - 2 x - 4 + v = 0$, i.e. $v = 5 - (x-1)^2$. This makes it clear that the range of values of your objective is $(-\infty, 5]$: there is a maximum but no minimum.

Answer 2

Given any $x$, you can calculate $y$ and $z$ from the constraints: $$z=x+2y-6\\x^2+y+(x+2y-6)=4\\3y=10-x-x^2\\y=\frac13(10-x-x^2)\\z=x+\frac23(10-x-x^2)-6$$ This is important, since there is a pair of $(y,z)$ for any $x$, and therefore for any $x$ you get $$y+z=4-x^2$$ Then $$2x+y+z=2x+4-x^2$$ It's easy to see that this expression has a maximum for $x=1$, with the corresponding $y=8/3$ and $z=1/3$, but there is no minimum value when $x\to\pm\infty$.