--- title: "Duality LPP min:Z=3x₁+2x₂" subject: "Mathematics III" chapter: 5 key: "semester_3/mathematics_iii:duality-lpp-min-z-3x-2x" source: https://www.bctengineer.com/q/semester_3/mathematics_iii/duality-lpp-min-z-3x-2x site: BCT Engineering --- # Duality LPP min:Z=3x₁+2x₂ Mathematics III · Chapter 5: Linear Programming - Asked in: 2080 Bhadra, 7 marks - Video: [Duality Problem 1,2 - Linear Programming Problems (LPP) - Engineering Mathematics - 4](https://www.youtube.com/watch?v=drU7dOa836M) — Ekeeda - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/mathematics_iii/duality-lpp-min-z-3x-2x - Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-5 ## Question Use the simplex method to solve the linear programming problem (constructing duality): Minimize Z = 3x₁ + 2x₂ subject to 3x₁ − x₂ ≥ −5, −x₁ + 4x₂ ≥ 1, x₁ + 9x₂ ≥ 6 and x₁, x₂ ≥ 0. ## Short answer By constructing the dual maximization problem and solving it with the simplex method, the optimal primal solution is x1 = 0, x2 = 2/3, giving minimum Z = 4/3. ## Answer **Step 1: Write the dual.** For the primal (minimisation, all constraints ≥), the dual is: ```text Primal: Minimize Z = 3x₁ + 2x₂ subject to 3x₁ − x₂ ≥ −5 −x₁ + 4x₂ ≥ 1 x₁ + 9x₂ ≥ 6 x₁, x₂ ≥ 0 Dual: Maximize W = −5y₁ + y₂ + 6y₃ subject to 3y₁ − y₂ + y₃ ≤ 3 −y₁ + 4y₂ + 9y₃ ≤ 2 y₁, y₂, y₃ ≥ 0 ``` Here y₁, y₂, y₃ are the dual variables, one for each primal constraint. **Step 2: Standard form of the dual.** Add slack variables s₁, s₂ (one per dual constraint). Since all right sides are positive, the slack variables give the starting basis. ```text Maximize W = −5y₁ + y₂ + 6y₃ subject to 3y₁ − y₂ + y₃ + s₁ = 3 −y₁ + 4y₂ + 9y₃ + s₂ = 2 ``` **Step 3: Simplex tables for the dual** **Table 1** ```text Cj -5 1 6 0 0 Basis CB y₁ y₂ y₃ s₁ s₂ b Ratio ----------------------------------- s₁ 0 3 -1 1 1 0 3 3 s₂ 0 -1 4 9 0 1 2 2/9 ----------------------------------- Zj 0 0 0 0 0 0 Cj-Zj -5 1 6 0 0 Entering: y₃ Leaving: s₂ Pivot = 9 ``` **Table 2** ```text Cj -5 1 6 0 0 Basis CB y₁ y₂ y₃ s₁ s₂ b ---------------------------------------- s₁ 0 28/9 -13/9 0 1 -1/9 25/9 y₃ 6 -1/9 4/9 1 0 1/9 2/9 ---------------------------------------- Zj -2/3 8/3 6 0 2/3 4/3 Cj-Zj -13/3 -5/3 0 0 -2/3 ``` **Step 4: Read the answer.** All Cj − Zj ≤ 0, so the dual is solved: y₁ = 0, y₂ = 0, y₃ = 2/9, Max W = 4/3. By the duality theorem, Min Z = Max W. The values of the primal variables are the numbers (without sign) in the Cj − Zj row under the slack columns: x₁ = 0 (under s₁), x₂ = 2/3 (under s₂). Check: Z = 3(0) + 2(2/3) = 4/3. **Answer:** x₁ = 0, x₂ = 2/3, Minimum Z = 4/3. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For a minimisation primal with all constraints >=, what form does the dual take? A. Minimise with >= constraints B. Maximise with <= constraints, one dual variable per primal constraint C. Maximise with >= constraints D. Minimise with <= constraints, one variable per primal variable Answer: B. The dual of a min problem with >= rows is a max problem with <= rows, with one dual variable for each primal constraint. (Key point: Forming the dual) 2. What is the dual objective for this problem? A. Maximize W = 3y1 + 2y2 B. Maximize W = 3y1 + 2y2 + 0y3 C. Maximize W = -5y1 + y2 + 6y3 D. Minimize W = -5y1 + y2 + 6y3 Answer: C. The right-hand sides (-5, 1, 6) of the primal constraints become the objective coefficients of the dual. (Key point: Dual objective) 3. How are the primal variable values obtained from the final dual table? A. From the Cj - Zj row under the slack columns (without sign) B. From the b column directly C. By solving the primal graphically D. From the Zj row under the y columns Answer: A. By the duality theorem the primal variables appear in the Cj - Zj row under the dual's slack columns. (Key point: Reading the primal solution) 4. What are the optimal value and primal solution? A. Min Z = 6, x1 = 2, x2 = 0 B. Min Z = 2/9, x1 = 0, x2 = 1 C. Min Z = 4/3, x1 = 2/3, x2 = 0 D. Min Z = 4/3, x1 = 0, x2 = 2/3 Answer: D. Max W = 4/3 equals Min Z, with x1 = 0 and x2 = 2/3; check 3(0) + 2(2/3) = 4/3. 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