1. **Stating the problem:** We are given a primal linear programming (LP) problem (P): $$\min \{-x_1 + 2x_2 - 38x_3\}$$ subject to $$3x_1 - 4x_2 + 5x_3 \leq -6$$ $$-7x_1 + 8x_2 - 9x_3 \geq 10$$ $$11x_1 - 12x_2 + 13x_3 = 14$$ with variable bounds: $$x_1 \geq 0, \quad x_2 \leq 0, \quad x_3 \text{ free}$$ 2. **Formula and rules for dual problem:** The dual problem is derived by associating a dual variable to each primal constraint: - For \(\leq\) constraints, dual variables are \(\geq 0\). - For \(\geq\) constraints, dual variables are \(\leq 0\). - For equality constraints, dual variables are free. The dual objective coefficients come from the right-hand side of primal constraints. The dual constraints come from the primal objective coefficients and variable bounds. 3. **Deriving the dual variables:** - Let \(u_1\) correspond to the first constraint (\(\leq\)) so \(u_1 \geq 0\). - Let \(u_2\) correspond to the second constraint (\(\geq\)) so \(u_2 \leq 0\). - Let \(u_3\) correspond to the third constraint (\(=\)) so \(u_3\) is free. 4. **Dual objective function:** Maximize $$-6u_1 + 10u_2 + 14u_3$$ 5. **Dual constraints:** From primal variables and their bounds: - For \(x_1 \geq 0\), dual constraint is: $$3u_1 - 7u_2 + 11u_3 \leq -1$$ (from primal objective coefficient \(-1\)) - For \(x_2 \leq 0\), dual constraint is: $$-4u_1 + 8u_2 - 12u_3 \geq 2$$ (from primal objective coefficient \(2\)) - For \(x_3\) free, dual constraint is equality: $$5u_1 - 9u_2 + 13u_3 = -38$$ (from primal objective coefficient \(-38\)) 6. **Answer to (a):** The dual problem has 3 variables \(u_1, u_2, u_3\) with: - \(u_1 \geq 0\) - \(u_2 \leq 0\) - \(u_3\) free So the correct choice is: **3 variables, 1 of them is free**. 7. **Answer to (b):** Given problem (D') matches exactly the dual problem derived above: - Objective: maximize \(-6u_1 + 10u_2 + 14u_3\) - Constraints and signs match - Variable bounds match Therefore, (D') is a correct dual of (P). **Final answers:** (a) 3 variables, 1 of them is free (b) yes