--- title: "Simplex method max:P=50x₁+80x₂" subject: "Mathematics III" chapter: 5 key: "semester_3/mathematics_iii:simplex-method-max-p-50x-80x" source: https://www.bctengineer.com/q/semester_3/mathematics_iii/simplex-method-max-p-50x-80x site: BCT Engineering --- # Simplex method max:P=50x₁+80x₂ Mathematics III · Chapter 5: Linear Programming - Asked in: 2078 Kartik, 7 marks; 2073 Chaitra, 10 marks; 2072 Chaitra, 10 marks; 2069 Chaitra, 10 marks (asked 4 times) - Video: [Simplex Method, Example 1](https://www.youtube.com/watch?v=rzRZLGD_aeE) — Dr D’s Math Help - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/mathematics_iii/simplex-method-max-p-50x-80x - Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-5 ## Question Use the simplex method to solve the following LPP: Maximize P = 50x₁ + 80x₂ subject to x₁ + 2x₂ ≤ 32, 3x₁ + 4x₂ ≤ 84, x₁, x₂ ≥ 0. ## Short answer Add slacks s1, s2; initial basis s1=32, s2=84. Simplex yields x1=20, x2=6, max P=1480. ## Answer **Step 1: Standard form.** Every constraint is of the ≤ type, so add one slack variable to each (s₁, s₂ ≥ 0) to turn it into an equation. Slack variables have zero cost. ```text Maximize P = 50x₁ + 80x₂ subject to x₁ + 2x₂ + s₁ = 32 3x₁ + 4x₂ + s₂ = 84 ``` **Step 2: Initial basic feasible solution.** Put x₁, x₂ = 0, so the slack variables are the basic variables: s₁ = 32, s₂ = 84. Then P = 0. Rules used: the entering variable is the one with the largest positive Cj − Zj; the leaving variable is the one with the smallest non-negative ratio b/column; the solution is optimal when every Cj − Zj ≤ 0. **Step 3: Simplex tables** **Table 1** ```text Cj 50 80 0 0 Basis CB x₁ x₂ s₁ s₂ b Ratio --------------------------------- s₁ 0 1 2 1 0 32 16 s₂ 0 3 4 0 1 84 21 --------------------------------- Zj 0 0 0 0 0 Cj-Zj 50 80 0 0 Entering: x₂ Leaving: s₁ Pivot = 2 ``` **Table 2** ```text Cj 50 80 0 0 Basis CB x₁ x₂ s₁ s₂ b Ratio ------------------------------------- x₂ 80 1/2 1 1/2 0 16 32 s₂ 0 1 0 -2 1 20 20 ------------------------------------- Zj 40 80 40 0 1280 Cj-Zj 10 0 -40 0 Entering: x₁ Leaving: s₂ Pivot = 1 ``` **Table 3** ```text Cj 50 80 0 0 Basis CB x₁ x₂ s₁ s₂ b -------------------------------- x₂ 80 0 1 3/2 -1/2 6 x₁ 50 1 0 -2 1 20 -------------------------------- Zj 50 80 20 10 1480 Cj-Zj 0 0 -20 -10 ``` **Step 4: Optimality test.** In the last table every Cj − Zj is ≤ 0, so the solution is optimal. **Answer:** x₁ = 20, x₂ = 6, and Maximum P = 1480. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which variable enters first? A. x2, with the largest Cj - Zj (80) B. s1 C. s2 D. x1, since 50 is smaller Answer: A. 80 > 50. (Key point: Entering variable) 2. In Table 1, which row leaves? A. s2, ratio 21 B. s1, ratio 16 beats 21 C. s1, ratio 32 D. s2, ratio 84 Answer: B. 32/2 = 16 and 84/4 = 21. (Key point: Ratio test) 3. When is the solution optimal? A. When b is zero B. When every Cj - Zj >= 0 C. When every Cj - Zj <= 0 D. When all slacks leave Answer: C. No variable can then increase the objective. (Key point: Optimality) 4. What is the optimum? A. x1 = 20, x2 = 6, P = 1480 B. x1 = 0, x2 = 16, P = 1280 C. x1 = 20, x2 = 6, P = 1280 D. x1 = 6, x2 = 20, P = 1900 Answer: A. 50(20) + 80(6) = 1480. 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