Linearization of a max function I have to cope with a constraint of the form (1) in the following problem: 
$$\begin{align}\max\quad& x+y\\
\text{s.t.}\quad&
x + y \leq \max \{x,y\} &(1)\\  
&0 \leq x \leq U_x&(2)\\ 
&0 \leq y \leq U_y&(3)\\
\end{align}$$
In the following link you can find an approach but I don't understand it.
https://www.leandro-coelho.com/how-to-linearize-max-min-and-abs-functions/
I don't understand: what is $S^+$, $S^-$ and how would a penalization look like? 
(I refer to the text: "The max function can be linearized as follows: ..." in the reference).
I would be grateful if somebody could help. 

The linked figure shows the problem in LP Format and the solution.
 A: The reformulation in the link isn't guaranteed to work. In this case, it doesn't, because the feasible region is not convex. You cannot express a non-convex feasible region with linear constraints.
To see that it is not convex, note that if $x\geq y$, then $x+y\leq x$, so $y=0$. Otherwise, $y>x$, so $x+y\leq y$ and then $x=0$. Therefore, either $x=0$ or $y=0$.
We have feasible solutions of $(x,y)=(U_x,0)$ and $(x,y)=(0,U_y)$. If the feasible region were convex, a convex combination of those would be feasible, like $(x,y)=\frac{1}{2}(U_x,U_y)$. However, it's not feasible unless $U_x=0$ or $U_y=0$.
So for general $U_x,U_y$, you can't use this reformulation. Instead, you'll probably want to introduce a binary variable, and solve a mixed integer programme. Or, in this case, by inspection.
A: Hello I hope it is not too late; the penalty could be representative like this:

max x + y
x <=s1 + s2
y <=s1 + s2
s1 >= x - y
s2 >= y - x
s1 - bigMb <= 0
s2 - bigM(1-b) <= 0
0 <= x <= Ux
0 <= y <= Uy

where b is binary variable
