Exercise 6.3.3

In problem 2, we have $c=\left[\begin{array}{c}\hfill 5\hfill \\ \hfill 3\hfill \\ \hfill 8\hfill \end{array}\right]$, $A=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 2\hfill \end{array}\right]$, $b=4$. Since there’s only 1 constraint in the primal problem, we have 1 variable $y$ in the dual problem, i.e. $y$ is a scalar here. So we need:

Where we have $A^{T}y=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 2\hfill \end{array}\right]y\le \left[\begin{array}{c}\hfill 5\hfill \\ \hfill 3\hfill \\ \hfill 8\hfill \end{array}\right]$, we conclude $y\le 3$, the maximum of $y^{T}b=\mathit{yb}=4y$ is $4\times 3=12$. This agrees with problem 2 where it finds its minimial of 12.

This demonstrates the duality of the linear programming.