Exercise with absolute maximum and minimum Find the absolute maximum and minimum of the function $\ f(x,y,z)= \ x^2+\ y^2+\ z^2 $ on the set $ \{(x,y,z): x^2=z^2+1 \}$. 
My solution and question: To find the critical points I used the formula
$ L(x,y,z,\lambda)=f(x,y,z)+\lambda(g(x,y,z))$ 
which in my case turns out to be 
$L(x,y,z,\lambda)=x^2+y^2+z^2+\lambda(x^2-z^2-1)$ hence deriving
\begin{equation}
\begin{cases}
2x+2x\lambda=0 \\
2y=0 \\
2z-2z\lambda=0 \\
x^2-z^2-1=0
\end{cases}
\end{equation}
doing all the steps you get to have  \begin{equation}
\begin{cases}
\lambda=-1\\
y=0\\
z=0\\
x=\pm 1
\end{cases}
\end{equation}
So I get critical points $\;(-1,0,0)\;$ e $\; (1,0,0)\;$ 
going to replace in the function $\;f(x,y,z)\;$ we get for both points the value $1.$
So are both points of maximum?And what is the minimum?
 A: As an alternative
$$f(x,y,z)= \ x^2+\ y^2+\ z^2 \land x^2=z^2+1 \implies g(y,z)=y^2+2z^2+1$$
and


*

*$g_y=2y=0$

*$g_z=4z=0$


therefore according to your derivation critical points are $(\pm 1,0,0)$.
To conclude note that since
$$g(y,z)=y^2+2z^2+1\ge 1$$
and
$$g(y,z)=1 \iff y=z=0$$
both points are point of global minimum.
A: $f(x,y,z)=1+2z^{2}+y^{2}\geq 1$ and $f(x,y,z)=1$ when $y=z=0$ and $x=1$ so the minimum value is $1$. You are not asked to find the critical points, so this is a complete answer. The function is not bounded so sup is $\infty$.
A: To classify if it is a max/min point, calculate the bordered Hessian matrix:
$$\bar{H}=\begin{vmatrix}0&\phi_x&\phi_y&\phi_z\\ \phi_x&L_{xx}&L_{xy}&L_{xz}\\ \phi_y&L_{yx}&L_{yy}&L_{yz}\\
\phi_z&L_{zx}&L_{zy}&L_{zz}\end{vmatrix}=\begin{vmatrix}0&2x&0&-2z\\ 2x&2+\lambda&0&0\\ 0&0&0&2\\
-2z&0&2&2-2\lambda\end{vmatrix},$$
where $\phi(x,y,z)=x^2-z^2-1$.
Minors:
$$\begin{align}\bar{H}_1&=\begin{vmatrix}0&2x\\ 2x&2+\lambda\end{vmatrix}=-4x^2<0; \\
\bar{H}_2&=0;\\
\bar{H}_3&=-16x^2<0.\end{align}$$
This is the condition for minimum. Hence, the found points are local minimum points.
A: To determine whether it is the minimum or maximum, there is a systematic method, but very complicated. However for your question, we note that when $|z|$ is sufficiently large, so is $|x|$ then $f$. Therefore $f$ is unbounded above, i.e. has no maximum. Then both of your point only could be of minimum. Since 
$$
f(x,y,z) = y^2 +2z^2 +1\geqslant 1
$$
on the set, and clearly $f(\pm 1,0,0)=1$, thus the points are minimum points. 
