Area of a lattice polygon in terms of its width 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
 A: $\alpha=3/8$ is sharp according to, say this 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.]
A: This solution is not true, sorry.
I can prove that $area(M)\geq 3d^2/8$.
First of all find $v\in \mathbb Z^2$ such that $w_v(M)$ is minimal.  Than by an affine transform $a$ put $v$ into vector $(1,1)$.
Now draw lines which give us widths of $a(M)$ in the directions $(1,0),(0,1)$. Now $a(M)$ is inside a horizontal stripe of width at least $d$, inside a vertical stripe of width at least $d$ and inside a diagonal (in the direction $(1,-1)$) stripe of width exactly $d/{\sqrt{2}}$.
Here I use usual widths.
Now look at the picture - we see a 6-gon (or 3,4-gon) $B$ as the intersection of the strips, we know that on each side of $B$ there is a vertex of $a(M)$. By playing with vertices one can find all the extremal cases (so, vertices of $a(M)$ must coincide with vertices of $B$).
Finally, in all the extremal cases the area of $a(M)$ is no less than $3d^2/8$ 
