Maxima, minima and inexistence of gradient of $f(x,y) = min\{f_1(x,y), f_2(x,y)\}$ Consider:
$f_1(x,y) = -27x^2+27x−27y^2 −54y−31,$
$f_2(x,y) = −27x^2 −54x −27y^2 −54y −58$
How to determine the set of points $(x,y)$ where the gradient of
$f(x,y) = min\{f_1(x,y), f_2(x,y)\}$
does not exist?
Also: how to find local (and global) maxima and minima of $f(x,y)$?
What I´ve been able to do so far?
I supposed (just not sure how to prove it) that at the intersection of $f_1$ and $f_2$ the gradient will not exist. So I found the point $(-\frac{1}{3},y)$ to be a parabola in $R^3$ where the gradient of $f(x,y)$ will not exist.
Regarding maxima and minima, how do I apply the rule of finding where the gradient equals zero in order to later analyse the Hessian matrix? I don´t have a single expression for $f(x,y)$. I´m confused about it.
 A: We can improve the way the function $f$ is written by the following expression:
$$\min\{f_{1},f_{2}\}=\frac{f_{1}+f_{2}}{2}-\frac{|f_{1}-f_{2}|}{2}$$Then,
\begin{align*}
f: \mathbb{R}^{2}&\longrightarrow \mathbb{R},\\
(x,y) &\longmapsto f(x,y)=\frac{-54x^{2}-27x-54y^{2}-108y-89}{2}-\frac{|81x+27 |}{2}
\end{align*}
By definition $\nabla f: \Omega\subseteq \mathbb{R}^{2}\longrightarrow \mathbb{R}^{2}$ over on open set $\Omega$ is defined as $(f_{1},f_{1})\longmapsto \left(\frac{\partial f_{1}}{\partial x}, \frac{\partial f_{2}}{\partial y} \right)$. Then,

*

*$\displaystyle \frac{\partial f_{1}}{\partial x}(x,y)=-\frac{27}{2}\left( 4x+1+\frac{81(3x+1)}{|81x+27|}\right)$.


*$\displaystyle \frac{\partial f_{2}}{\partial y}(x,y)=-54(y+1)$.
So $${\rm Dom} \nabla f=\{(x,y)\in \mathbb{R}^{2}: 81x+27\not=0\}$$
and this answer your first question about the domain of gradient of $f$.
About your another question for the local minimum and local maximum you can use the theorem stated in the answer here: https://math.stackexchange.com/a/4405749/1027216.
Finally about another question for the global minimum and global maximum you need more information about the domain of $f$ at least check in the problem if there is no additional information about a domain other than all of $\Bbb{R}^2$. Because for example setting $x=0$, we get the function does not have lower bound. Also for the critical points you need more information about the domain for $f$.
