State the dual problem for the following, but do not solve it. Minimize $w=y_{1}+4 y_{2}+6 y_{3}+9 y_{4}$ Subject to: \[ \begin{aligned} 3 y_{1}+6 y_{2}+8 y_{3}+y_{4} & \geq-3 \\ y_{1}+6 y_{2}+6 y_{3}+y_{4} & \geq 15 \\ y_{1} \geq 0, y_{2} \geq 0, y_{3} \geq 0, y_{4} & \geq 0 \end{aligned} \] \[ \text { Maximize } z=\square x_{1}+\square x_{2} \] Subject to: $\leq 1$ $\leq 4$ $\leq 6$ $\leq 9$ $x_{1} \geq 0, x_{2} \geq 0$
Solution
Step 1 ：The dual problem of a linear programming problem is another linear programming problem that is derived from the original problem. The dual problem is obtained by interchanging the roles of the objective function and the right-hand side constants, and by interchanging the roles of the decision variables and the constraints. In this case, the original problem is a minimization problem with four constraints. Therefore, the dual problem will be a maximization problem with four decision variables. The coefficients of the objective function in the dual problem will be the right-hand side constants of the original problem, and the coefficients of the constraints in the dual problem will be the coefficients of the objective function in the original problem.
Step 2 ：The dual problem is: Maximize \(z=-3x_{1}+15x_{2}\) subject to: \[\begin{aligned} 3x_{1}+x_{2} & \leq 1 \\ 6x_{1}+6x_{2} & \leq 4 \\ 8x_{1}+6x_{2} & \leq 6 \\ x_{1}+x_{2} & \leq 9 \\ x_{1} \geq 0, x_{2} \geq 0, x_{3} \geq 0, x_{4} & \geq 0 \end{aligned}\]
Step 3 ：Final Answer: \(\boxed{\text{Maximize } z=-3x_{1}+15x_{2} \text{ subject to: } \begin{aligned} 3x_{1}+x_{2} & \leq 1 \\ 6x_{1}+6x_{2} & \leq 4 \\ 8x_{1}+6x_{2} & \leq 6 \\ x_{1}+x_{2} & \leq 9 \\ x_{1} \geq 0, x_{2} \geq 0, x_{3} \geq 0, x_{4} & \geq 0 \end{aligned}}\)
