I have a polyhedron:
$$\begin{align} P = \{x \in R^3 \;\mid\;& x_1 + 2x_2 \ge 4 \tag1 \\ &2x_1 − x_2 \ge 3 \tag2 \\ &x_2 \ge 1 \tag3 \\ &x_1 + x_3 \ge 4 \tag4 \\ &2x_1 + 2x_2 − x_3 \ge 7 \tag5 \\ &x_1, x_2, x_3 \ge 0 \tag6 \}\end{align}$$
- Determine the face F induced by $I = \{(3),(4)\}$.
- Determine an inequality $c x \leq \gamma$, such that $c x \leq \gamma$ is valid for $P$ and $F = \{x \in P \mid c x = b\}$.
- Determine the dimension of $F$.
- Prove that inequality (5) is valid but does not define a supporting hyperplane
I have never seen this type of problem so I have no idea how would I need to find the face of this or what does even the polyhedron mean written like this which are edges?
If anyone can help it would be amazing
What you have here is the typical way a polyhedron (or, more generally, a polytope) is described in linear optimization: as intersection of half-spaces.
Each inequation is a half-space. Some will define facets (they are "facet-defining") but some may not, if the intersection of all other half-spaces is already contained in the interior of this half-space.
I assume here the question is about the face induced by (3) and (4) considered as equations, i.e. with a "$=$" instead of a "$\lt$". And also that in (3) we have $x_2$, not $x^2$. Then the face is the intersection of two planes, so a line. The part of it included into $P$ may be an edge, or empty, if one of the inequations is not facet-defining. To know that, you have to report $x_2=1$ and $x_1+x_3=4$ in the other inequations, and check if there are solutions. (I have no pen and paper right now, but it seems to be the case).
We can assume $c$ is a vector, $\gamma$ a real number, but $b$? In usual linear programming notation (see Wikipedia page linked above), $b$ is the vector of right-hand constants in all inequations; so a $8$-dimension vector (there are $8$ inequations); but that would make $cx=b$ a nonsense. So $b$ is probably a real number. But still the question is clumsy, because there is no quantifier for $b$, and moreover, asking for "$cx\le \gamma$ such that it is valid $\forall x \in P$" would have been sufficient. Anyway, for a given $c$, $\gamma$ has to be the $\max$ of all $b$ such that $F$ is not empty. It can be $\infty$, if the polyhedron is not finite (it may be "open" on one side - I have not checked).
F is the intersection of a plane and a polyhedron, so a polygon (which may be infinite if the polyhedron is infinite), or empty. So when not empty, it has dimension $2$.
I assume "a supporting hyperplane" means a facet-supporting hyperplane, i.e. the inequation is facet-defining. So the goal is to find a linear combination of inequations $(1)$ to $(4)$, and the $3$ inequations in $(6)$, with positive or null coefficients, that implies inequation $(5)$. However it is not possible: point $(2,1,2)$ verifies all inequations except $(5)$. So there is an error, in the text or in the way I understand it.
Note that, in a typical optimization problem, variables and inequations are counted in thousands, and cases with millions of variables and inequations are not rare. So knowing which inequations are redundant is really a hard problem.