find extrema of $f(x,y,z)=z$ with domain $D=\{(y^2+z^2)/6\le x,x^2+y^2=z^2+16\}$ in $\mathbb R$
on $f:D \to R ,f(x,y,z)=z$ 


*

*First off just with a brief look at my function I can say that there are no critical points ($f_x=0,f_y=0,f_z=1$), (morover its an open set which means that the max and min cannot occur ? correct me if I'm wrong)

*I can now analyze the domain $D$, what I can say is that :


$(y^2+z^2)/6\le x\Longrightarrow$ It's a kind of Paraboloid
$x^2+y^2=z^2+16 \Longrightarrow$ It's a Hyperboloid of One Sheet
If I want to find the boundary of $D$, I need to put those two equations in a system, finding: $x^2-2z^2+6x-16=0$ which is the intersection between those two surfaces.


*

*Now in order to find possible min/max points, I can use the Lagrange multiplier system with  $x^2-2z^2+6x-16=0$ as a constraint.


$$\lambda (2x+6)=0$$
$$0=0$$
$$1+\lambda (-2z)=0$$
$$x^2-2z^2+6x-16=0$$


*

*From the first equation I can say that it is TRUE for $\lambda = 0$ or $x=-3$ , but $\lambda$ cannot be zero becouse It doesn't satisfy the third equation . What I can do instead is using $x=-3$ in the forth equation , but here is the problem : I remain with $-z^2=25$ and I conclude that I didn't find any points. Even if I use the rhird equation finding $z$ and putting it inside the forth equation it still gives me something like $\lambda^2 $  equal to a negative number.

*Where did I make the mistake? (maybe the boundary ?)
 A: Using a slack variable $\epsilon$ to convert the inequality into an equality we have
$$
L(z,y,z,\lambda,\mu,\epsilon) = z+\lambda\left(\frac 16(y^2+z^2)-x+\epsilon^2\right)+\mu(x^2+y^2-z^2-16)
$$
the stationary points are the solutions for
$$
\nabla L = \left\{
\begin{array}{rcl}
 2 \mu  x-\lambda &=&0 \\
 \frac{\lambda  y}{3}+2 \mu  y&=&0 \\
 \frac{\lambda  z}{3}-2 \mu  z+1&=&0 \\
 \epsilon ^2-x+\frac{1}{6} \left(y^2+z^2\right)&=&0 \\
 x^2+y^2-z^2-16&=&0 \\
 2 \epsilon  \lambda &=&0 \\
\end{array}
\right.
$$
which are
$$
\left[
\begin{array}{cccccc}
x & y & z & \lambda & \mu & \epsilon \\
 8 & 0 & -4 \sqrt{3} & \frac{2 \sqrt{3}}{5} & \frac{\sqrt{3}}{40} & 0 \\
 8 & 0 & 4 \sqrt{3} & -\frac{2 \sqrt{3}}{5} & -\frac{\sqrt{3}}{40} & 0 \\
\end{array}
\right]
$$
As we can observe the two stationary points are with $\epsilon = 0$ meaning that the restriction is actuating and they are at the feasible region border.
Attached a plot showing in red the extrema locations as well as the intersection curve between the borders of the two restrictions in blue.

Attached the MATHEMATICA script to produce the plot

h = (y^2 + z^2)/6 - x;
g = (x^2 + y^2 - z^2 - 16);
p1 = {8, 0, -4 Sqrt[3]};
p2 = {8, 0, 4 Sqrt[3]};
gr1 = Graphics3D[{Red, Sphere[p1, 0.3]}];
gr2 = Graphics3D[{Red, Sphere[p2, 0.3]}];
gr0 = ContourPlot3D[{h == 0, g == 0}, {x, -10, 10}, {y, -10, 
    10}, {z, -10, 10}, 
   MeshFunctions -> {Function[{x, y, z, f}, h - g]}, 
   MeshStyle -> {{Thick, Blue}}, Mesh -> {{0}}, 
   ContourStyle -> 
    Directive[Orange, Opacity[0.5], Specularity[White, 30]], 
   PlotPoints -> 40];
Show[gr0, gr1, gr2, PlotRange -> All]
`
A: The domain $D$ is a two-dimensional wobbly disc cut out from the hyperboloid. The boundary of this disc is the curve $\gamma$ of intersection between the hyperboloid and the paraboloid. This curve is determined by two constraints (and not by one, as you seem to believe).
Drawing a figure showing the projection to the $(x,z)$-plane suggests that the point on $\gamma$ with maximal $z$-coordinate is on the symmetry plane $y=0$. There are two such points, namely $P_\pm:=\bigl(8,0, \pm4\sqrt{3}\bigr)$. These two points would then lead to the extremal values $\pm4\sqrt{3}$ of $f\restriction D$.
Doing it the hard way we first remark that  the hyperboloid has no horizontal tangent planes. It follows that necessarily the extrema of $f\restriction D$ are taken on $\partial D=\gamma$. We therefore set up the Lagrangian
$$\Phi:=z-\lambda(6x-y^2-z^2)-\mu(x^2+y^2-z^2-16)$$
and then obtain the equations
$$\eqalign{\Phi_x&=-6\lambda-2\mu x=0 \cr
\Phi_y&=2\lambda y-2\mu y=0\cr \Phi_z&=1+2\lambda z+2\mu z=0\ .\cr}$$
From the second equation it follows that $y=0$ or $\lambda=\mu$. The case $y=0$ leads to the points $P_\pm$  we have already found. When $\lambda=\mu$ we are left with the equations
$$ \lambda(3+x)=0,\quad 4\lambda z=-1\ .$$
This enforces $\lambda\ne0$, hence $x=-3$. But there are no points in $D$ with $x=-3$.
To sum it up: The extrema of $f\restriction D$ are taken in the points $P_\pm$.
