Find all of the extreme values using Lagrange Multipliers I need to find all of the extreme values of the function
$x^2+y^2+z^2$
constrained to $x^2+2y^2-z^2-1=0$
the problem is that I get this system that I have no idea how to solve:
$2x=2\lambda x$
$2y=4\lambda y$
$2z=-2\lambda z$
$x^2+2y^2-z^2-1=0$
All I can think of is that $x=y=z=0$ but then in the last equation I would get $-1=0$ which of course can't happen.
Any idea what do I need to do here?

Define,

*

*$f(x,y,z)=x^{2}+y^{2}+z^{2}$ and notice that $f\in C^{+\infty}$ so $f\in C^{1}$.

*$g(x,y,z)=x^{2}+2y^{2}-z^{2}$ and notice that $g\in C^{+\infty}$ so $g\in C^{1}$.

*$k=1$.

So Lagrange multipliers says $$\begin{cases} \nabla f=\lambda g,\\ g=k\end{cases} \implies \begin{cases}2x=2\lambda x,\\ 2y=4\lambda y,\\2z=-2\lambda z,\\x^{2}+2y^{2}-z^{2}=1 \end{cases} \implies \begin{cases}x=\lambda x,\quad (1)\\ y=2\lambda y,\quad (2)\\ z=-\lambda z,\quad (3)\\x^{2}+2y^{2}-z^{2}=1,\quad (4) \end{cases}$$
Now, we are looking for candidates for the extremes

*

*In $(1)$ if $\boxed{\lambda=1}$ so by $(2)$ we need $y=0$ and by $(3)$ we need $z=0$ and by $(4)$ we need $x^{2}=1$. Hence, here we have the candidates $\{(1,0,0),(-1,0,0)\}$.


*In $(2)$ if $\boxed{\lambda=1/2}$ so by $(1)$ we need $x=0$ and by $(3)$ we need $z=0$ and by $(4)$ we ned $2y^{2}=1$. Hence we have the candidates $\{(0,\sqrt{2}/2,0),(0,-\sqrt{2}/2,0)\}$.


*In $(3)$ if $\boxed{\lambda=-1}$ so by $(2)$ we need $y=0$ and by $(1)$ we need $x=0$ and by $(4)$ we need $-z^{2}=1$ but so $z^{2}=-1$ absurd.
All the candidates are given by
$$\{(1,0,0),(-1,0,0),(0,\sqrt{2}/2,0),(0,-\sqrt{2}/2,0)\}$$
Now,

*

*$f(1,0,0)=1$

*$f(-1,0,0)=1$.

*$f(0,\sqrt{2}/2,0)=\left(\frac{\sqrt{2}}{2}\right)^{2}=\frac{1}{2}$.

*$f(0,-\sqrt{2}/2,0)=\left(-\frac{\sqrt{2}}{2}\right)^{2}=\frac{1}{2}$.

Well this is the interesting part in this problem because at this point one could say "the minimum is reached at the points $(0,\pm \sqrt{2}/2,0)$ and the maximum is reached at the points $(0,\pm 1,0)$" which is the usual kind of conclusions when one uses Lagrange multipliers. However, we cannot say that here. The natural question is "why can't we say that?" Well the short answer is that when we use the Lagrange multiplier theorem we are assuming the
existence of a smooth function in order to optimise (we have that to $f$ ), smooth constraints (we have that to $g$), with region described by the closed and bounded constraint and this is the problem. To see this, look at the graph of the constraint region below:
$\hspace{4cm}$ 
So, this allows us to rule out the existence of a maximum just by looking at the contours of the graph and we can think of the existence of the minimum also just by looking at the contour of the constraint region and considering the graph for $f$, see below.
$\hspace{4cm}$ 
Then since the geometrical interpretation of the Lagrange multipliers is that: the extreme values of $f$ correspond to the level surfaces touching $g=1$ then,

*

*We can state that there is no global maximum,


*And by other hand the minimum are given at $(0,- \sqrt{2}/2,0)$ with value minimum $1/2$.
Now if you want an analytical approach you can use the Hessian's test for optimization with the Langragian function  $\mathscr{L}(x,y,z,\lambda)=x^{2}+y^{2}+z^{2}+\lambda(x^{2}+2y^{2}-z^{2}-1)$.
