First. It's easier to understand the problem by describing the application where it arises from.
We have a convex body $B$ in $\mathbb{R}^{3}$ and measure points on its surface. The measurements are noisy. We need to reconstruct the body from these measurements. It's dangerous to do this just by considering $Hull(x_{1}, \ldots, x_{N})$, because not all points are vertexes of $Hull(x_{1}, \ldots, x_{N})$.
Now we state it as a mathematical problem.
Suppose that N points $x_{1}, \ldots, x_{N}$ in $\mathbb{R}^{3}$ are given (and they are so that $O = (0, 0, 0)$ is the interior point of $Hull(O, x_{1}, \ldots, x_{N})$).
Let's consider the set $\mathfrak{M}$ of all closed convex surfaces $M$ that have the following property: $O = \{0, 0, 0\} \in Int \overline{M}$, where $\overline{M}$ is a convex body whose bondary is $M$, i. e. $\partial \overline{M} = M$
It's needed to find such $M \in \mathfrak{M}$ that minimizes the sum of distances from $x_{i}$ to $M$ for all $x_{1}, \ldots, x_{N}$.
$$ I(M) = \sum \limits_{i = 1}^{N} \rho(x_{i}, M) \to inf \;\;\; s.t. M \in \mathfrak{M} $$
It's obvious that it is enough to consider not whole $\mathfrak{M}$, but only polyhedrons, because if $M^{*} = argmin \; I(M)$ then one can construct a polyhedron $P^{*}$ such that $I(M^{*}) = I(P^{*})$ just by finding such $y_{i} \in M^{*}$ so that $\rho(x_{i}, M^{*}) = \rho(x_{i}, y_{i})$ and then $P^{*} = Hull(y_{1}, \ldots, y_{N})$.
Does anybody know any method to find such best surface or polyhedron?
NOTE: It is not a convex hull. Here is the contrary instance:
If you consider the convex hull, then it's possible to find better surface by dropping 3 exterior points and considering the convex hull of the rest points. It will have better value of the inital functional.