Find the extrema of the implicit function $f(x,y,z(x,y)) = x^2 + y^2 - z^2$
Of course, I start with calculating the partial derivatives by implicit differentiation.
$$\frac{\partial z}{\partial x} = \frac x z$$
$$\frac{\partial z}{\partial y} = \frac y z$$
Which yields that the only feasible stationary point is $(0,0)$.
By the general formula of the function, $z = 0 $.
But now we have a problem. Since $z = 0$, no partial derivatives at this point exist.
We could try to check if $(0,0,0)$ is an extremum straight from the definition, but we don't have the formula for $z$.
I need to find the extremum of $z$.
How do I proceed form here?
You know that : $$ x^2 + y^2 = z^2 $$ $$\frac{\partial z}{\partial x} = \frac x z$$ $$\frac{\partial z}{\partial y} = \frac y z$$
We have that :
$$ \nabla z = ( \frac {x}{ \sqrt{x^2 + y2}} , \frac {y}{ \sqrt{x^2 + y2}}) $$
So apart from $0$, $\nabla z \neq 0$. Thus, on $\mathbb R^*$, there is no critical point.
Moreover, You have that
$ O := (0,0,0) $,
$ E := \{ x,y,z \,| x^2 + y^2 = z^2 \} $,
$$ O \in E $$
Finally, if we take a look at a point $$ A = (\epsilon,\epsilon,\sqrt 2 \epsilon) $$ for a given, small $\epsilon$, we find that $$ A \in E$$ and $A_z = \sqrt 2 \epsilon > 0$.
We have proven that the origin is a minimum of the function, since you can get as close as you wish from $0$ and you will always be higher than the origin: $A_z > 0$.
So yes, your function is $ \mathcal C^1( \mathbb R^2 / \{(0,0)\})$ thus you can't derivate at $0$, but you can still proove it is a maximum, because the function is continuons on $\mathbb R^2$.