TITLE: Optimum solution to a Linear programming problem
QUESTION [17 upvotes]: If we have a feasible space for a given LPP (linear programming problem), how is it that its optimum solution lies on one of the corner points of the graphical solution? (I am here concerned only with those LPP's which have a graphical solution with more than one corner/end point.)
I was asked to take this as a lemma in the class, but got curious about the proof. Any help is sincerely appreciated.

REPLY [9 votes]: Jiri's answer gives the intuitive explanation.  Formally, the fact that an optimal solution lies at an extreme point is a consequence of the representation theorem for polyhedra and the fact that the feasible region of a linear program is a polyhedron.  
The representation theorem says that a polyhedron is the convex sum of its vertices plus the nonnegative linear combination of its extreme directions.    More formally, if $S$ is a polyhedron, $\{{\bf v}_1, {\bf v}_2, \ldots, {\bf v}_k\}$ is the set of extreme points of $S$, and $\{{\bf d}_1, {\bf d}_2, \ldots, {\bf d}_m\}$ is the set of extreme directions of $S$ then ${\bf x} \in S$ if and only if ${\bf x} = \sum_{i=1}^k \alpha_i {\bf v}_i + \sum_{i=1}^m \beta_i {\bf d}_i$, $\sum_{i=1}^k \alpha_i = 1$ and $\alpha_i, \beta_i \geq 0$.
Then we have
Theorem: If $S$ is the feasible region of some linear program with objective function $z = {\bf c}^T {\bf x}$ then 1) $z$ attains its optimal value at some vertex of $S$, 2) the linear program is infeasible, or 3) the linear program is unbounded. 
Proof: First, assume, without loss of generality, that the LP wants to maximize $z$.  If the LP is infeasible, then $S$ is empty.  Otherwise, there is at least one point ${\bf x} \in S$.  If $S$ has an unbounded extreme direction ${\bf d}_i$ such that ${\bf c}^T {\bf d}_i > 0$, then the LP is unbounded, as we can make the value of ${\bf c}^T \left(\sum_{i=1}^k \alpha_i {\bf v}_i + \beta_i {\bf d}_i\right)$ as large as we wish by increasing $\beta_i$.  Assume, then, that ${\bf c}^T {\bf d}_i \leq 0$ for each $i$.  Let ${\bf v}^*$ be the vertex with the largest objective function value.  Then, for ${\bf x} \in S$, $${\bf c}^T {\bf x} = {\bf c}^T \left(\sum_{i=1}^k \alpha_i {\bf v}_i + \sum_{i=1}^m \beta_i {\bf d}_i \right) = \sum_{i=1}^k \alpha_i ({\bf c}^T {\bf v}_i) + \sum_{i=1}^m \beta_i ({\bf c}^T {\bf d}_i) \leq \sum_{i=1}^k \alpha_i ({\bf c}^T {\bf v}^*) = {\bf c}^T {\bf v}^*.$$  Thus $z$ must attain its optimal value at ${\bf v}^*$.