Solution
= Solution
\b[No.] For the <linear polyhedron> $P=[0,1]^2$, take $c=(1,0)^T$ and $x^*=(1,1/2)^T$. The maximum of $c^Tx$ is one, attained at $x^*$. Yet
$$
x^*=\tfrac12(1,0)^T+\tfrac12(1,1)^T,
$$
so $x^*$ is not an <extreme point>, hence not a <vertex of a linear polyhedron>. An <optimal face> can contain many <points>; an arbitrary maximizer need not be one of its <vertices of a linear polyhedron>.