Let $M$ be a lattice polygon on a plane (i.e. its vertices are integer points $(i,j)\in\mathbb Z^2$).


Let us define lattice width in a direction $v=(m,n)\in\mathbb Z^2$ as $w\_v(M)=\max\limits\_{x,y\in M} v\cdot(x-y)$.


Suppose the $minimal$ lattice width of $M$ equals $d$. It is clear that the area of $M$ should be proportional to $d^2$.


One can prove an inequalities $area\geq d^2/4$ as is written in comments.
But it seems far from the best one.


So, the question is what is the best $\alpha$ such that $area(M)\geq \alpha d^2$ for each $M$ with minimal lattice width equals $d$. From my comment below one can extract that $\alpha\leq 3/8$


Added: F. Petrov gave a link to the proof of the fact, but there is no complete proof in the Internet so I rewrote it in the appendix in <http://arxiv.org/abs/1306.4688>



Area of a lattice polygon in terms of its width
$\alpha=3/8$ is sharp according to, say [this](http://www.math.illinois.edu/~z-furedi/PUBS/furedi_barany_local-diam.pdf) article, the authors refer to 
[L. Fejes-Toth and E. Makai, Jr., On the thinnest non-separable lattice of convex
plates, Studia Sci. Math. Hungar. 9 (1974), 191–193.]


