One very promising place to look is at [Paley Graphs][1] which are self dual (so $\alpha=\omega$.) This [answer][2] to a question suggests that, based on prime $n \lt 10000,$ it might be that $\alpha \omega =O(log^4 n).$ Although it been the subject of a fair amount of research, all that is known for sure is that $\log n \lt \alpha=\omega \le \sqrt{n}.$ The upper bound can, evidently, be reduced to $\sqrt{n}-1$ infinitely often.

Any other circulant graphs could be considered and those with degree $\frac{n-1}{2}$ (so edge density $\frac12$) do seem optimal. A pentagon is a circulant graph (in fact a Paley Graph) and the tensor product of circulant graphs is a circulant graph. 

Any graph of $R(3,3)=6$ vertices or more has $\max(\alpha,\omega) \ge 3.$ In general, one less than a Ramsey number $R(s,s)-1$ is the largest size for a graph with $\max(\alpha,\omega) \lt s.$

$R(4,4)-1=17$ and among all graphs on 17 vertices (regular or not) the Paley Graph on $17$ vertices is the unique example with $\alpha \lt 4$ and $\omega \lt 4.$

No larger Ramsey numbers are known although $42 \lt R(5,5) \le 49.$ No graph on $42$ vertices could be a Paley graph and there are at least $656$ known with $\alpha=\omega=4.$ The Paley Graph on $37$ vertices has $\alpha=\omega=4$ but the Paley Graph on $41$ vertices has $\alpha=\omega=5.$ 

So Paley graphs are perhaps not the absolute best but they are easily described explicit (self dual, vertex transitive) graphs that seem quite good. As far as I know, the largest know graph with $\alpha \lt 6$ and $\omega \lt 6$ is the Paley Graph on $101$ vertices.

A final quote:

> Paley graphs are quasi-random ([Chung et al. 1989][3]): the number of times
> each possible constant-order graph occurs as a subgraph of a Paley
> graph is (in the limit for large q) the same as for random graphs, and
> large sets of vertices have approximately the same number of edges as
> they would in random graphs.

**LATER** It occurs to me that a random circulant graph with edge density $\frac12$ might be a better choice in some ways. On one hand the chance of $\{{1,2,3\cdots,k\}}$ being a clique in that model is $\frac1{2^k}$ but, due to the multiplicative nature of quadratic residues, in the Paley graph this just requires the primes up to $k$ to be residues hence it is $\frac1{2^{\ln k}}=\frac1{k^{\ln2}}.$ On the other hand, if we somehow knew that the expected value of $\alpha \omega$ is less than $1000 \log n^2,$ that wouldn't give us any explicit graph.


  [1]: https://en.wikipedia.org/wiki/Paley_graph
  [2]: http://mathoverflow.net/a/107979/8008
  [3]: https://dx.doi.org/10.1007%2FBF02125347