We have the set $$M=\{(x,y,z)\in\mathbb R^3: x^2 + y^2 = z \wedge x+y+z=12\}$$ and the function $$F(x,y,z) = xy+ z^2.$$
How can we find the global maximum and global minimum of F on M and prove their existence?
I tried to use Lagrange multipliers:
$$G(x,y,z)=F(x,y,z)-\lambda_1 m_1(x,y,z)-\lambda_2 m_2(x,y,z)$$
$$G(x,y,z)=xy+ z^2 -\lambda_1 (\ x^2 + y^2-z)-\lambda_2(x+y+z-12)$$
and then I counted the first partial derivatives of $G(x,y,z)$ by $x$, $y$, and $z$.
I don't know how to continue from this point.
If you consider that the locus of the constraints $ \ M \ $ is the intersection of an "upward-opening" circular paraboloid and an oblique plane, said plane being symmetrical about the plane $ \ y \ = \ x \ $ and sloping "downward" in the "positive" direction, then the constraint curve is a tilted ellipse with its "high end" in the third quadrant of the $ \ xy- $ plane and its "low end" in the first quadrant. Since the function $ \ F (x, \ y, \ z) \ $ has symmetry about the same plane, we might already guess that the global maximum lies at the high end of the ellipse and that since the product $ \ xy \ $ is negative in the second and fourth quadrants, the global minimum is likely be somewhere on the ellipse in each of those quadrants (again by symmetry of the curve and the function).
So saying, the two-constraint form of the "Lagrange equations" is
$$ y \ = \ 2 \ \lambda \ x \ + \ \mu \ \ , \ \ x \ = \ 2 \ \lambda \ y \ + \ \mu \ \ , \ \ 2 \ z \ = \ -\lambda \ + \ \mu \ \ . $$
We can immediately "eliminate" $ \ \mu \ $ by writing the single equation
$$ \mu \ \ = \ \ y \ - \ 2 \ \lambda \ x \ \ = \ \ x \ - \ 2 \ \lambda \ y \ \ = \ \ 2 \ z \ + \ \lambda \ \ . $$
(It is the symmetries mentioned above that are permitting us to do this. The algebra encountered in extremization problems isn't always this obliging.)
We can re-arrange the first pair of terms as
$$ x \ ( \ 1 \ + \ 2 \ \lambda \ ) \ \ = \ \ y \ ( \ 1 \ + \ 2 \ \lambda \ ) \ \ \Rightarrow \ \ ( \ x \ - \ y \ ) \ ( \ 1 \ + \ 2 \ \lambda \ ) \ \ = \ \ 0 \ \ . $$
So we see that same $ \ x \ - \ y \ $ factor cropping up that Kf-Sansoo shows, which means we want to check values of $ \ F \ $ at the ends of the ellipse where $ \ x \ = \ y \ $ . (As to where that is, use this in the constraints and solve the system $ \ 2 \ x^2 \ = \ z \ \ , \ \ 2 \ x \ = \ 12 \ - \ z \ $ . )
As to the other factor, which gives $ \ \lambda \ = \ - \frac{1}{2} \ $ , we need to see where it goes. Now we require $ \ x \ \neq \ y \ $ in our " $ \mu $ " equation:
$$ y \ - \ 2 \ \left( - \frac{1}{2} \right) \ x \ \ = \ \ x \ - \ 2 \left( - \frac{1}{2} \right) \ y \ \ = \ \ 2 \ z \ + \ \left( - \frac{1}{2} \right) \ \ , $$
part of which is redundant. We now find $ \ x \ + \ y \ \ = \ \ 2 \ z \ - \frac{1}{2} \ $ , which can be inserted into the second (linear) constraint equation to find $ \ z \ $ , and these results can be used with the first constraint to find $ \ x \ $ and $ \ y \ $ (we will find two points for this value of $ \ z \ $ , lying in the plane $ \ y \ = \ -x \ $ , again as expected from the geometrical situation.)
[ADDENDUM: In checking things over, I went all the way through to the end and found that there are no solutions under this second condition -- check for yourself why this is so. Thus the two points found for $ \ x \ = \ y \ $ are the global extrema.]