Is it true that if LP $ \text{max} \{c^Tx \ | \ Ax \leq b \}$ has an optimal solution, then $\exists$ a vertex which is simultaneously an optimal solution for LP?
I know this works for LP of a standard type, i.e. $ \text{max} \{c^Tx \ | \ Ax = b \}$, but does it works for the canonical form of the LP?
Yes. Wlog. $c\ne 0$. If $x_0$ is an optimal solution, then the hyperplane $c^Tx=c^Tx_0$ intersects the solution space in a convex polyhedron. This polyhedron has a vertex $x_1$ (if if the polygon happens to be unbounded). If $x_1$ is not a vertex of the original polyhedron, then there is a line segment through $x_1$ that has points on both "sides" of the hyperplane $c^Tx=c^Tx_0$, but the points on one of the sides have a better value of the target function.