This content is copy of comment of @Kushagra gupta
Maximum :-
number of edges in a $bipartite\ graph$ with $n$ vertices $=\left \lfloor{\dfrac{n^2}{4}}\right\rfloor$
Proof:
- Let on one side $k$ vertices & then on the other side there will be $(n-k)$ vertices (assumed)
- Total number of edges $=k(n-k)$
Let's maximise this value:
$\\\frac{\partial }{\partial x}(kn-k^2)=0 \\\\n-2k=0 \\\\ \dfrac{n}{2}=k$
$\frac{\partial^2 }{\partial x^2}(kn-k^2)\\ \\ =-2<0$
means at $k=\dfrac{n}{2}$ maximum edges will be present.
$Ans: =\left \lfloor{\dfrac{12^{2}}{4}}\right\rfloor=36$