\[ \begin{array}{ll} \text { Minimize } & C=11 x_{1}+2 x_{2} \\ \text { subject to } & 4 x_{1}+x_{2} \geq 25 \\ & 3 x_{1}+x_{2} \geq 6 \\ & x_{1}, x_{2} \geq 0 \end{array} \] a. Form the dual problem. \[ \begin{array}{ll} \text { Maximize } & P=\square \mathrm{y}_{1}+6 \mathrm{y}_{2} \\ \text { subject to } & \square \mathrm{y}_{1}+\square \mathrm{y}_{2} \leq 11 \\ & \mathrm{y}_{1}+\square \mathrm{y}_{2} \leq \square \\ & \mathrm{y}_{1}, \mathrm{y}_{2} \geq 0 \end{array} \]
Solution
Step 1 ：Form the dual problem of the given linear programming problem. The objective function of the dual problem is formed from the right-hand side of the original problem's constraints, and the constraints of the dual problem are formed from the coefficients of the original problem's objective function. The signs of the inequalities are also reversed.
Step 2 ：The coefficients of the objective function in the original problem are 11 and 2, and the right-hand side of the constraints are 25 and 6. Therefore, the objective function of the dual problem should be \(25y_1 + 6y_2\).
Step 3 ：The coefficients of the constraints in the original problem are 4, 1, 3, and 1. Therefore, the constraints of the dual problem should be \(4y_1 + 3y_2 \leq 11\) and \(y_1 + y_2 \leq 2\).
Step 4 ：Finally, the dual problem is: \[\begin{array}{ll} \text { Maximize } & P=25 y_{1}+6 y_{2} \\ \text { subject to } & 4 y_{1}+3 y_{2} \leq 11 \\ & y_{1}+y_{2} \leq 2 \\ & y_{1}, y_{2} \geq 0 \end{array}\]
