XbenX wrote: What is the maximal number of solutions can the equation have $$\max \{a_1x+b_1, a_2x+b_2, \ldots, a_{10}x+b_{10}\}=0$$where $a_1,b_1, a_2, b_2, \ldots , a_{10},b_{10}$ are real numbers, not all equal to $0$. I guess that there is some condition missing, or that the problem is misstated. In the current formulation, you may choose $a_1=a_2=\cdots=a_{10}=0$, $b_1=b_2=\cdots=b_9=-1$ and $b_{10}=0$. Then the left hand side is identically 0, and all real $x$ are solutions.

@above, fixed, and thanks for pointing it out.

XbenX wrote: What is the maximal number of solutions can the equation have $$\max \{a_1x+b_1, a_2x+b_2, \ldots, a_{10}x+b_{10}\}=0$$where $a_1,b_1, a_2, b_2, \ldots , a_{10},b_{10}$ are real numbers, not all $a_i$ equal to $0$. In this new formulation, we may choose $a_1=b_1=0$ (so that $a_1x+b_1\equiv0$) and $(a_i,b_i)=(-1,0)$ for $i=2,\ldots,n$. Then all $x\ge0$ are solutions of $\max \{a_1x+b_1, a_2x+b_2, \ldots, a_{10}x+b_{10}\}=0$.

Right condition is all $a_i$ are not equal $0$

RagvaloD wrote: Right condition is all $a_i$ are not equal $0$ Aha. Then the maximum number is 2. Two solutions are possible: Consider the situation with $a_1=1$, $a_2=a_3=\cdots=a_{10}=-1$ and $b_1=\cdots=b_{10}=-1$. Then the only two solutions of $0=\max \{a_1x+b_1, a_2x+b_2, \ldots, a_{10}x+b_{10}\}=\max\{x-1,-x-1\}$ are $x=1$ and $x=-1$. More than two solutions is impossible: The maximum of all these linear functions is a convex function $f$ If a convex function $f$ takes the value $0$ at three different places $x_1<x_2<x_3$, the function is identically $0$ on the interval $[x_1,x_3]$; but then one of the lines $a_ix+b_i$ must be identically $0$, which implies $a_i=0$.

Obvious, all solutions are in form $x_i=-\frac{b_i}{a_i}$ Let $f_i(x)=a_ix+b_i$ Let $x_1 \neq x_2$ are solutions ,then $f_1(x_1)=0,f_2(x_1) \leq 0 \to \frac{-a_2b_1+a_1b_2}{a_1} \leq 0$ and $f_2(x_2)=0, f_1(x_2)=\frac{-a_1b_2+a_2b_1}{a_2} \leq 0$ $0 \leq f_1(x_2)f_2(x_1)= \frac{-(a_1b_2-a_2b_1)^2}{a_1a_2} \to a_1a_2<0$ So $x_i \neq x_1,x_i \neq x_2$ can not be solution because both inequalities $a_ia_1<0$ and $a_ia_2<0$ ca not be true at same time

Wow this problem is cool