\[ \begin{array}{ll} \text { Minimize } C=x_{1}+9 x_{2} \\ \text { subject to } & 5 x_{1}+2 x_{2} \geq 1 \\ & 6 x_{1}+5 x_{2} \geq 6 \\ & x_{1}, x_{2} \geq 0 \end{array} \] a. Form the dual problem. \[ \begin{array}{ll} \text { Maximize } & P=\square y_{1}+\square y_{2} \\ \text { subject to } & \square y_{1}+\square y_{2} \leq \square \\ & \square y_{1}+\square y_{2} \leq \square \\ & y_{1}, y_{2} \geq 0 \end{array} \]

Step 1 ：Form the dual problem of the given linear programming problem. The dual problem is another linear programming problem derived from the original problem. The objective function of the dual problem is to maximize the minimum value of the original problem. The constraints of the dual problem are derived from the coefficients of the original problem.

Step 2 ：In this case, the dual problem of the given problem can be formed by taking the coefficients of the constraints as the coefficients of the objective function and the coefficients of the objective function as the right-hand side of the constraints. The inequalities are also reversed.

Step 3 ：Finally, the dual problem is: \[\begin{array}{ll} \text { Maximize } & P=y_{1}+6 y_{2} \\ \text { subject to } & 5 y_{1}+6 y_{2} \leq 1 \\ & 2 y_{1}+5 y_{2} \leq 9 \\ & y_{1}, y_{2} \geq 0 \end{array}\]

Step 4 ：\(\boxed{\begin{array}{ll} \text { Maximize } & P=y_{1}+6 y_{2} \\ \text { subject to } & 5 y_{1}+6 y_{2} \leq 1 \\ & 2 y_{1}+5 y_{2} \leq 9 \\ & y_{1}, y_{2} \geq 0 \end{array}}\)