Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 37 / 2 / a / Solution

The feasible point $x=(8/5,0,1/5)$ makes the second and third constraints tight; the first has left side two. Its objective is $27/5$.

To prove optimality from first principles, multiply each of the second and third inequalities by $3/5$ and add. For every nonnegative feasible $x$,This is an explicit weak duality bound, proved here simply by adding inequalities. The displayed point attains it, soEquality forces $x_2=0$ and both contributing constraints tight, proving uniqueness as well.