Proof for the maximal number of edges in an unconnected graph

Question

I was asked to show that a graph $G$ with 11 vertices and 46 edges must be connected. I'd like to start with claiming that if $G$ were not connected, the maximal number of edges would be obtained with one component that has 10 vertices (And forms a clique) and the last vertex alone. This is obviously true, but how does one justify such a claim? (Preferably for the general case)

Answer 1

Hint: Suppose it is not connected. Then we have at least two nonempty subgraphs $A$ and $B$ so that we can't reach from $A$ any point in $B$ and vice versa. Say $|A| = a>0$ and $|B| =b>0$. Then we have at most $${a\choose 2}+{b\choose 2}$$ edges where $a+b=11$. Can you finish?

Let $a=n$, so $b=11-n$. We make a quadratic function $$f(n) = {n\choose 2}+{11-n\choose 2} = n^2-11n+55$$ with $n\in \{1,2,...10\}$ and you have to prove that $f(n)<46$.

Answer 2

I think you can justify your method by considering that the subgraph, say $G_1$, with $10$ vertices has at most $45$ edges when $G_1 = K_{10}$ (Notice that $45$ comes from Handshaking Lemma when each vertex has degree $9$). And since $G_2$ has $1$ vertex and no edge, the result follows. In general, Aqua's solution is really well thought.