 Solution

The dual from part (a) isThe point $(y,z)=(1,2)$ is dual feasible and has value $29$. The primal pointis feasible and also has objective valueBy weak duality, neither point can be improved, soThe 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.