Let $A\in\Bbb{R}^{m\times n},\ b\in\Bbb{R}^m, c\in\Bbb{R}^n.$ Consider the linear program:$$\;\;\;\;\qquad \max c^Tx\\ \text{s.t.}\quad Ax\leq b$$Assume that there is a feasible $w$, with $c^Tw\gt 0$ and $Aw\leq0$. Show that the LP is unbounded.
For a polyhedron $\mathcal{P}=\{x\mid Ax\leq b\}$, setting $b=0$: $\;\;\mathcal{P}(A,0)$ describes a cone. Further if the set of solutions of $\mathcal{P}(A,0)$ contains more than just the origin $\{0\}$, meaning $\mathcal{P}(A,0)\neq \{0\}$, it is unbounded. It is given, that for a point $w$ in $\mathcal{P}(A,0)$, $\;\; c^Tw\gt 0$, therefore $\mathcal{P}(A,0)\neq \{0\}$ holds and we can conclude that the LP is unbounded.
I am not sure that my attempt is sufficient for the proof. I would very much appreciate it, if anybody were to point out any mistakes or provide a hint or a solution to this.
Thank you
A couple of points: first, it is not immediately obvious how $\mathcal{P}(A,0)$ relates to the original feasible region $\mathcal{P}$. Moreover, it is possible for an LP to be bounded when $\mathcal{P}$ is not. For example, $\min_{x \geq 0} 1^Tx$ is a bounded LP, but $\{x \geq 0\}$ is not.
The crucial observation is that, for any $x \in \mathcal{P}$ and $\lambda \geq 0$, one has $$ A(x + \lambda w) = Ax + \lambda A w\leq b+0=b, $$ so $x + \lambda w \in \mathcal{P}$. What's more, $c^T(x+\lambda w) = c^Tx+\lambda c^Tw$. This means that $c^T(x + \lambda w) \to \infty$ as $\lambda \to \infty$, which tells us that the LP is unbounded.