 Solution

Suppose the unique optimum $x^{\ast }$ were $(1-t)u+tv$ for distinct feasible points $u,v$ and $0<t<1$. Feasibility is preserved by convex combinations. Optimality gives $c^{T}u\le c^T{x}^{\ast }$ and $c^{T}v\le c^T{x}^{\ast }$, while linearity gives their strictly weighted average equal to $c^T{x}^{\ast }$. Both inequalities must therefore be equalities. Then $u$ and $v$ are also optimal, contradicting uniqueness. Hence the unique optimum is an extreme point of the feasible set.