Simple optimization problem - finding the critical points

Question

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?

Answer 1

1. 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.

2. 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.

3. 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.