# How to determine both the primal and dual solution for a min problem?

Consider the minimization problem

$$\min 5x_1 + 2x_2 + 4x_3$$ subject to \begin{align} 3x_1 + x_2 + 2x_3 & \geq 4 \\ 6x_1 + 3x_2 + 5x_3 & \geq 10 \\ x_1, x_2, x_3 & \geq 0 \end{align} My question: I have found the dual problem to the minimazation that being $$\max 4y_1 + 10y_2$$ subject to \begin{align} 3y_1 + 6y_2 & \leq 5 \\ y_1 + 3y_2 & \leq 2 \\ 2y_1 + 5y_2 & \leq 4 \\ y_1,y_2 & \geq 0 \end{align} and have found the optimal solution $$y* = (y_1*,y_2*) = (1,1/3)$$.

However, now I am asked to specify an optimal solution for both the primal and the dual problem but I am not sure how to? I tried to solve the minimization problem using the simplex method, but I just couldn't figure it out. All the videos, I have watched, said to do it by solving the dual problem.. What do I do?

• You can solve them both by the simplex method. Also, the primal has three decision variables, so your solution is incomplete. Commented Oct 19, 2020 at 17:25
• I meant that $y* = (1,1/3)$. Isn't that correct? Commented Oct 19, 2020 at 17:26
• I think I figured it out.. Commented Oct 19, 2020 at 17:38
• @Mathias What have you figured out? Commented Oct 20, 2020 at 14:06

You can apply complementary slackness to obtain $$x_3=0$$ (because $$2y_1^*+5y_2^*<4$$) and $$3x_1 + x_2 + 2x_3 = 4$$ and $$6x_1 + 3x_2 + 5x_3 = 10$$ (because $$y_1^* > 0$$ and $$y_2^* > 0$$).