I am reading the book "Engineering Optimization: Theory and Practice, 5th Edition" on Chapter 7 and page 409 and stumble upon this system of equation
$\begin{array}{*{20}{c}} {\left\{ {\begin{array}{*{20}{c}} {{{\left( {{x_1} + 1} \right)}^2} - 2r\left[ {\max \left( {0,1 - {x_1}} \right)} \right] = 0}\\ {1 - 2r\left[ {\max \left( {0,1 - {x_1}} \right)} \right] = 0} \end{array}} \right.}&{\left( 1 \right)}\\ {\left\{ {\begin{array}{*{20}{c}} {\min \left[ {{{\left( {{x_1} + 1} \right)}^2},{{\left( {{x_1} + 1} \right)}^2} - 2r\left( {1 - {x_1}} \right)} \right] = 0}\\ {\min \left[ {1,1 + 2r{x_2}} \right] = 0} \end{array}} \right.}&{\left( 2 \right)} \end{array}$
Also $r>0$
I do not know how can they get from (1) to (2) so I think that there is some max-min arithmetic behind it. Therefore, my question is:
What is the underlying max-min arithmetic that transform (1) to (2) ?
Thank you for your enthusiasm !
Here are the two basic properties you need : $$ \max(x,y) + a = \max(x+a,y+a) $$ and $$ c \cdot \max(x,y) = \begin{cases} \max(cx,cy) & \mathrm{if\;} c \ge 0 \\ \min(cx,cy) & \mathrm{if\;} c \le 0 \end{cases} $$ In the present case, you have $a = (x_1 + 1)^2$ and $c = -2r$ with $r > 0$.