Simple optimization problem - finding the critical points Let $f(x,y,z):=x+y+2z$ with the set $G:=\{(x,y,z)\in\mathbb R^3 | z^2 = 1 + x^2 + y^2, z\geq 0\}$
Get the relative local extrema of $f$ on $G$. Decide if the found points are maxima or minima.
My Calculations:
Looking at $\mathring{G}$
$df(x,y,z)=(1,1,2)\overset{!}{=}0 \Rightarrow $ no critical Points
Looking at $\partial G$
define $g(x,y,z):=x^2+y^2-z^2+1=0$
Define Lagrangian: $L:=(x,y,z,\lambda)=f-\lambda g = x+y+2z-\lambda(x^2+y^2-z^2 + 1)$
Not that $\lambda\neq 0$
$\frac{\partial L}{\partial x}=1-2\lambda x \overset{!}{=} 0 \Rightarrow x=\frac{1}{2\lambda}$
$\frac{\partial L}{\partial y}=1-2\lambda y \overset{!}{=} 0 \Rightarrow y=\frac{1}{2\lambda}$
$\frac{\partial L}{\partial z}=2+2\lambda z \overset{!}{=} 0 \Rightarrow z=\frac{-1}{\lambda}$
$\frac{\partial L}{\partial \lambda} = -g(x,y,z) \overset{!}{=} 0$
We plug $x,y,z$ into $g$ and get: $x^2+y^2-z^2+1=\frac{1}{4\lambda^2}+\frac{1}{4\lambda^2}-\frac{1}{\lambda^2}+1\overset{!}{=} 0 \Rightarrow \lambda = \pm\frac{1}{\sqrt{2}}$
For $\lambda = \frac{1}{\sqrt{2}}$ we get $x=\frac{1}{\sqrt{2}}, y=\frac{1}{\sqrt{2}}, z=-\sqrt{2}$
For $\lambda = -\frac{1}{\sqrt{2}}$ we get $x=\frac{-1}{\sqrt{2}}, y=\frac{-1}{\sqrt{2}}, z=\sqrt{2}$
Note that $z\geq 0$ so $z=-\sqrt{2}$ is not valid.
We get the following points which might be an extrema:
$p_1=(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},\sqrt{2})$
$p_2=(\frac{-1}{\sqrt{2}},\frac{1}{\sqrt{2}},\sqrt{2})$
$p_3=(\frac{1}{\sqrt{2}},\frac{-1}{\sqrt{2}},\sqrt{2})$
$p_4=(\frac{-1}{\sqrt{2}},\frac{-1}{\sqrt{2}},\sqrt{2})$
We evalute f in $p_i$ getting: $f(p_1)>f(p_2)=f(p_3)>f(p_4)$
So $p_1$ is maxima and $p_4$ is minima.
Question 1: I usually do the evaluating of the found ciritcal points with the hessian matrix of the lagrangian. But does the above (last step) work as well?
Question 2: Something here is wrong. The solution only finds $p_4$ as only critical point. But I can't see whats wrong with my calculation. All my points fullfill $g$.
Question 3: I'm unsure about the $\overset{!}{=}$ I used. E.g. here: $1-2\lambda x \overset{!}{=}$ is it needed? I usually use it if there's a variable I have to calculate but one might argue, that since it's only one variable in there, it is not needed. Im not sure.
Thanks
Edit: I think this is more a problem of understanding what I'm doing. Somehoe I should recognize, that only $p_4$ is a valid critical point - but how?
 A: *

*Between two maxima there has to be minima. However, I would stringly advise on using the Hessian instead of sorting function values. Especially when using multivariate functions, the meaning of "between" is not always obvious. 

*Since you said it yourself: $z<0$ ist invalid. That means, that $\lambda \neq -\frac{1}{\sqrt{2}}$!
It cannot be that value. If it would, you violate a constraint. 
By using this, you can not put this value into the formulas for $x$ and $y$, since this value of the Lagrange-Multiplier is not a valid solution. 
Additionally: Your points $p_2$ and $p_3$ use $\lambda>0$ for one value and $\lambda <0$ or another. But in a single evaluation of your Lagrangian, $\lambda$ must be constant. 
If you would have multiple $\lambda$ (by having multiple equality-constraints) you would have to "mix" them. But also then, you have to kepp in min, if your chioce is a valid one. 

*I use $\overset{!}{=}$ all the time, when solving equations. It does not matter, what kind of equation (univariate, linear, homogenous, etc). I thikn, it is properly used here. 
