This is not possible (yet or never). I will show that this problem is NP-complete by reducing SAT into your problem.
Consider an instance of SAT as a boolean formula under conjunctive normal form, using $c$ clauses and $n$ different variables ($v_1$, $v_2$,…, $v_n$). We will make an instance of your problem with vector size $n+1$ and $1\le i \le c$.
First let $A$ be the identity matrix of size $n+1$, and $b$ the vector with only $1$. Hence $Ax<b$ and $x>0$ implies that $x$ is a boolean vector (only $0$s and $1$s).
Next consider the clause $i$ formed by a disjunction of $v$ variables. Let $c_{ij}$ ($1\le j\le v$) such that :
- if the $j$th variable used in the clause $i$ is positive and equals to $v_m$, let $c_{ij}$ be a vector with only $0$ except in position $m$ equals to $1$.
- if the $j$th variable used in the clause $i$ is negative and equals to $\neg v_m$, let $c_{ij}$ be a vector with only $0$ except in position $m$ equals to $-1$ and $1$ in position $n+1$.
If you try to maximize $m_i=\max_{j=1}^{v}c_{ij}^Tx$, you remark that :
- you can put a $1$ at position $n+1$ in $x$, because in $c_{ij}$ at position $n+1$ you will always have $0$ or $1$.
- you need to have a $1$ at the right position if $c_{ij}$ is linked to a positive variable in clause $i$ to have $c_{ij}^Tx=1$, else $c_{ij}^Tx=0$.
- you need to have a $0$ at the right position if $c_{ij}$ is linked to a negative variable in clause $i$ to have $c_{ij}^Tx=1$, else $c_{ij}^Tx=0$.
Hence $m_i=1$ if and only if the clause $i$ is satisfied by $x$ (considering the $n$ first coordinates of $x$ as true/false value for variable $v_1$, …, $v_n$).
So, the formula is satisfiable iff the maximum of the instance created is equal to $c$ (the number of clauses satisfied).
Hence your problem is NP-hard, because this reduction is polynomial (quadratic indeed)
It is NP-complete, because with the right $x$, you can easily compute the maximum (and verify any assumption on it).
Note that you can fill with $c_{ij}=0$, if you need to have $1\le i\le k$ and $1\le j \le k$ with the same $k$ for both.