= Solution
{c}
At $\mathbf y^*=(4/3,2)$, the first three dual left sides are $10/3$, $2/3$ and $14/3$, strictly above their bounds $1$, $-1$ and $-5$. Thus <complementary slackness> forces $x_1=x_2=x_3=0$. Both <dual variables> are positive, so both primal inequalities are equalities:
$$
3x_4+3x_5=2,\qquad 4x_4+x_5=1.
$$
Solving this <linear system> gives
$$
\boxed{\mathbf x^*=\left(0,0,0,\frac19,\frac59\right),\qquad
\max\left(x_1-x_2-5x_3+12x_4+6x_5\right)=\frac{14}{3}.}
$$
All components are nonnegative, the two original upper bounds are met exactly, and the <objective function> equals the dual value. The <weak duality> bound therefore certifies optimality. The forced zero components and unique solution of the two remaining equalities also prove that the primal optimum is unique.
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