--- title: "Simplex method max:Z=15x₁+10x₂" subject: "Mathematics III" chapter: 5 key: "semester_3/mathematics_iii:simplex-method-max-z-15x-10x" source: https://www.bctengineer.com/q/semester_3/mathematics_iii/simplex-method-max-z-15x-10x site: BCT Engineering --- # Simplex method max:Z=15x₁+10x₂ Mathematics III · Chapter 5: Linear Programming - Asked in: 2072 Kartik, 10 marks; 2068 Baishakh, 10 marks (asked 2 times) - Video: [Simplex method Maximizez = 4x_1 + 10x_2Subject 2x1 + x2 = 502x1 + 5x2 = 100 2x1 + 3x2 = 90](https://www.youtube.com/watch?v=c8er-5I7q5g) — Engineering & 12(Sci) with Amey Sir - 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-z-15x-10x - Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-5 ## Question Use the simplex method to solve the linear programming problem: Maximize Z = 15x₁ + 10x₂ subject to 2x₁ + 2x₂ ≤ 10, x₁ + 3x₂ ≤ 10 and x₁, x₂ ≥ 0. ## Short answer Add slack variables s₁, s₂ to convert inequalities. Initial BFS (0,0,10,10). Pivot x₁ in, then all Cj-Zj≤0. Optimal: x₁=5, x₂=0, max Z=75. ## 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 Z = 15x₁ + 10x₂ subject to 2x₁ + 2x₂ + s₁ = 10 x₁ + 3x₂ + s₂ = 10 ``` **Step 2: Initial basic feasible solution.** Put x₁, x₂ = 0, so the slack variables are the basic variables: s₁ = 10, s₂ = 10. Then Z = 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 15 10 0 0 Basis CB x₁ x₂ s₁ s₂ b Ratio --------------------------------- s₁ 0 2 2 1 0 10 5 s₂ 0 1 3 0 1 10 10 --------------------------------- Zj 0 0 0 0 0 Cj-Zj 15 10 0 0 Entering: x₁ Leaving: s₁ Pivot = 2 ``` **Table 2** ```text Cj 15 10 0 0 Basis CB x₁ x₂ s₁ s₂ b ------------------------------ x₁ 15 1 1 1/2 0 5 s₂ 0 0 2 -1/2 1 5 ------------------------------ Zj 15 15 15/2 0 75 Cj-Zj 0 -5 -15/2 0 ``` **Step 4: Optimality test.** In the last table every Cj − Zj is ≤ 0, so the solution is optimal. **Answer:** x₁ = 5, x₂ = 0, and Maximum Z = 75. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What slack equations are formed? A. x1 + s1 = 10 B. 2x1 + 2x2 + s1 = 10 and x1 + 3x2 + s2 = 10 C. 2x1 + 2x2 - s1 = 10 D. 2x1 + 2x2 + s1 = 0 Answer: B. One slack per <= constraint. (Key point: Standard form) 2. In Table 1, why does s1 leave? A. s1 is artificial B. Largest ratio C. Smallest ratio, 10/2 = 5 D. It has the larger coefficient Answer: C. The ratio test picks the smallest non-negative ratio. (Key point: Pivot) 3. Why is Table 2 optimal? A. Z has reached 100 B. x2 is zero C. b is positive D. All Cj - Zj are <= 0 (0, -5, -15/2, 0) Answer: D. No entering variable improves Z. (Key point: Optimality) 4. What is the answer? A. x1 = 5, x2 = 0, Z = 50 B. x1 = 5, x2 = 0, Z = 75 C. x1 = 0, x2 = 5, Z = 50 D. x1 = 5, x2 = 5, Z = 125 Answer: B. 15(5) = 75. 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