How to prove point is vertex not interior point Say we have a polyhedron represented by:
  x1 + x2 + 2*x3 <=3 
3*x1 + x2 +   x3 <=4
  x1,  x2,    x3 >= 0

how can I demonstrate that (0,3,0) is a vertex and not an interior point?
to make these equations instead of inequalities, we can use slack variables:
  x1 + x2 + 2*x3 + s1 = 3 
3*x1 + x2 +   x3 + s2 = 4

my guess is that at least one slack variable has to be 0. But it could also be only one slack variable can zero...not sure yet.
 A: From the point of view of calculus, this is a solid bounded by planes (setting the three inequalities to equalities) in ℝ3. The edges are line segments in intersections of each two planes, $\textbf{the vertices the intersection of any two edges, i.e., the intersection of any three boundary planes.}$
All variables being nonnegative means that the other faces are given by $x=0,y=0,z=0$, or the $xy,yz,xz$-planes. And the polyhedron should be in the first octant (all variables nonnegative).
Note that $(0,3,0)$ lies in the three planes $x_1+x_2+2x_3=3, x=0, z=0$, so it must be a vertex.
A: In my estimation, to find a vertex, if you have N variables and M equations, at least (N-M) variables have to take on the value of 0.
this smells good:

*

*N-M vars are 0

*equations are satisfied

*all variables are non-negative

I believe (1) is necessary, and (2) is necessary, and given (3), then the total combination is sufficient for a point in a polyhedron to be a vertex.
(Without a restriction that all variables are +positive, it seems like there are pretty much always infinite solutions).
