Volume of convex lattice polytopes with one interior lattice point Let $P$ be a convex polytope in $\mathbb{R}^3$ whose every vertex lies in the $\mathbb{Z}^3$ lattice.
Question: If $P$ contains exactly one lattice point in its interior, what is the maximum possible volume of $P$? 
Notice that a convex lattice polytope with no interior lattice point can have arbitrarily large volume, but if the number of interior lattice points in it is a given finite number, then its volume cannot be arbitrarily large. In two dimensions the answer is known: A convex lattice polygon with exactly one interior point has volume at most $9/2$, which follows from a more general inequality by P.R. Scott (see On Convex Lattice Polygons, Bull. Austral. Math. Soc., vol 15; 1976, 395-399).
The best example I know in that respect is the tetrahedron with vertices $(0,0,0),\ (4,0,0),\ (0,4,0)$ and $(0,0,4)$, with volume ${32}\over3$. Is this perhaps the maximum volume? Of course, the same question can be asked in higher dimensions as well, with an analogous example to consider as a possible candidate for an answer.
In view of several examples presented in answers and comments below, it seems that the optimal polytope should be a simplex, in every dimension. Has this been conjectured already?
 A: This is addressed in the 2013 paper (appeared in Advances in 2015) by Averkov, Krumpelmann, Nill. The give a sharp bound for the volume of a lattice simplex with one interior lattice point (Theorem 2.2 in the paper), and an improved bound for a general lattice polyhedron with the same property (theorem 2.7) (the two bounds are not the same, indicating that Wlodek's conjecture is either still open or false). The results are stated in terms of the Sylvester sequence:
$$s_1 = 2; s_i = 1 + \prod_{j=1}^{i-1} s_j.$$
With that, the volume of tbe biggest simplex in $d$ dimension with one lattice point is bounded (with equality achieved) by
$$ \frac2{d!}(s_d-1)^2,$$
while for arbitrary polytopes, the bound is $$(s_{d+1}-1)^d,$$ so a lot worse.
A: Depictions of the two lattice polyhedra mentioned so far (by Wlodek Kuperberg & js21):

          


          

Left: $(0,0,0),(4,0,0),(0,4,0),(0,0,4)$. Right: $(0,0,0), (2,0,0), (0,3,0), (0,0,12)$.


A: Just to extend the examples so far: Given a non-decreasing list of $n$ integers $a_1 \leq a_2 \leq \cdots \leq a_n,$ consider the simplex with vertices at the origin and the points with all coordinates zero except $a_i$ in the $i$th position. The volume is $\frac{\prod a_i}{n!}.$ The point $(1,1,\cdots,1)$ is in the interior provided $\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_{n-1}}+ \frac{1}{a_n}\lt 1$ and there are no other lattice points in the interior provided  $\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_{n-1}}+ \frac{2}{a_n}\ge 1.$ Equality here puts the point $(1,1,\cdots,1,2)$ on the "sloped" face.
Subject to these constraints, I find that the optimal edge lengths with corresponding volume are the previously mentioned:
$3,3\rightarrow \frac{9}{2}$
$2,3,12 \rightarrow 12$ and $2,6,6\rightarrow 12$
$2, 3, 7, 84\rightarrow 147$
along with
$2, 3, 7, 43, 3612 \rightarrow 54360 \frac35$
$2,3,7,43,1807,6526884 \rightarrow 29583482464 \frac9{10}.$
These seem,with two small exceptions, to be given by,  $a_i=1+\prod_{j=1}^{i-1}a_j$ except that $a_n=2 \prod_{j=1}^{n-1}a_j.$
This puts the point $(1,1,\cdots,1,2)$ on the boundary.
A: this is a great question and fairly good answers are known for arbitrary dimensions. (I think there are works for $d=3$ but I am not sure.) The relevant papers are
On lattice polytopes having interior lattice points, by 
J.M. Wills; J. Zaks; M.A. Perles . 
Elemente der Mathematik (1982)
Volume: 37, page 44-45 (lower bounds)
and 
Bounds for lattice polytopes containing a fixed number of interior points 
by JC Lagarias, GM Ziegler - Canad. J. Math, 1991‏ (upper bounds)
Here are the lower bounds, $n$ is the number of points in the interior. ($v$ is the function.)

