= Solution
The dual from part (a) is
$$
\begin{aligned}
\text{maximise}\quad&7y+11z\\
\text{subject to}\quad
&y+z\leq3,\qquad y\leq2,\qquad z\leq2,\\
&y+2z\geq2,qquad y,z\geq0.
\end{aligned}
$$
The point $(y,z)=(1,2)$ is dual feasible and has value $29$. The primal point
$$
(x_1,x_2,x_3,x_4)=(7,0,4,0)
$$
is feasible and also has objective value
$$
3(7)+2(4)=29.
$$
By <weak duality>, neither point can be improved, so
$$
\boxed{\min=29,\qquad(x_1,x_2,x_3,x_4)=(7,0,4,0)}.
$$
The strict dual inequalities for $x_2$ and $x_4$, together with <complementary slackness>, force $x_2=x_4=0$ at any optimum; the two tight primal constraints then force $x_1=7$ and $x_3=4$. Thus the displayed minimizer is unique.
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