Solving min-max equations

Question

I'm struggling solving the following systems of equations: Solve for $x$ and $y$ $$ \min(2x,5y)+\min(3x,-2y)=-50 \\ \max(-3x,2y)+\max(6x,3y)=~~51 $$ I have tried many method however I'm still not getting the exact solution.

Answer 1

Hint:

Given that

$$\min(a,b)=a\text{ or }\min(a,b)=b$$ (and similarly for $\max$) you can try all the replacements of $\min/\max$ by one of their arguments, and this will generate $16$ linear systems of two equations in two unknowns. After resolution, you plug the values in the original equation and check validity.

Tedious, but not difficult.

First set:

$$\begin{align}5x=-50,\\2x-2y=-50,\\3x+5y=-50,\\3y=-50.\end{align}$$

Second set:

$$\begin{align}3x=51,\\-3x+3y=51,\\6x+2y=51,\\5y=51.\end{align}$$

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Update:

By changing the signs in the first equation,

$$\max(-2x,-5y)+\max(-3x,2y)=50$$

you reduce the number of cases from $16$ to $8$.

Answer 2

Let us consider "max" operator (the same can be said about $\min$ operator).

Between the 2 cases $\max(a,b)=a$ and $\max(a,b)=b$, a transition occurs when $a=b$.

But this transition on an equality condition geometrically means in our case the crossing of the boundary of some region which are straight lines with equations:

$$2x=5y, \ 3x=-2y \ (\text{twice}), \ 6x=3y$$

As they all pass through the origin, they generate "only" the $6$ "sectors" which can be seen on the figure with $3$ treated cases (up to you for the other ones):

In each region, one gets a system of 2 equations with 2 unknowns and have to solve it.

Caution : one must verify that the solution obtained by solving the system attached to a given sector belongs to this sector. !

This is precisely the issue with the blue sector case in which the $x$ solution is $x=-1/8$ : therefore solution $(x,y)$... isn't a solution because it doesn't belong to the given sector.