About the surface area vs. volume of polytopes Given a convex body $K\in\mathbb{R}^n$, represented by a set of linear inequalities (intersection of halfspaces), I am interested in understanding how much of its volume can be close to its perimeter (under certain restrictions).
More formally, given a parameter $k$, and a partition of $\mathbb{R}^n$ into boxes of side $k$, I would like to know how large can the ratio $|P|/|I|$ be, where $P$ is the set of such boxes which intersect the perimeter, and $I$ is the set of boxes fully contained in $K$ (such a bound would be important for determining the accuracy of integration, for example).
I believe that as in 2 and 3 dimensions, the smallest ratio would be achieved by a ball (sphere), and that the worst ratio would be achieved for polytopes whose volume approaches 0 (by having a width smaller than k in one dimension, for example, which gives $|I|=0$).
However - are there some reasonable limits (for example, containing the 1/n unit sphere, or just having a volume > 0, and a poly(n) representation length of the linear inequalities) that can determine an upper bound on this ratio?
Many thanks,
Guy
 A: A few updates to the question:


*

*Thanks for the correction about the minimality of a square (I was thinking of surface area/volume ratio, which is similar but different).

*$k$ is not a fixed parameter, but can vary with a given $K$. However, I cannot just let $k\rightarrow 0$, since I am thinking about an algorithm which is limited in space (and running time), and would like to understand how small does $k$ need to be in order to guarantee that the above ratio $|P|/|I|$ is not too bad (for example, that it is $\leq 1$).
So - to rephrase the above question: Let's assume that $K$ is a polytope in $\mathbb{R}^n$, and that it has a representation using poly(n) bits, as a set of linear inequalities.
 We also assume that it has volume greater than 0 (easy to verify). If it is helpful, we may even assume that it contains the $\frac{1}{n}$ sphere in $\mathbb{R}^n$. 
How small should the parameter $k$ (determining how fine we integrate) be, to determine that $|P|/|I|\leq 1$ ?
Thanks again,
Guy
