--- title: "Simplex method max:Z=3x₁+5x₂" subject: "Mathematics III" chapter: 5 key: "semester_3/mathematics_iii:simplex-method-max-z-3x-5x" source: https://www.bctengineer.com/q/semester_3/mathematics_iii/simplex-method-max-z-3x-5x site: BCT Engineering --- # Simplex method max:Z=3x₁+5x₂ Mathematics III · Chapter 5: Linear Programming - Asked in: 2083 Baishakh, 1+9 marks - Video: [Lec -5 Simplex Method || General form to Standard form || In Hindi || Easy steps to convert](https://www.youtube.com/watch?v=Y2T3MMvaEPU) — Start Practicing - 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-3x-5x - Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-5 ## Question Describe slack variables, and hence solve the following linear programming problem applying the simplex method: Maximize Z = 3x₁ + 5x₂ subject to 3x₁ + 2x₂ ≤ 18, x₁ ≤ 4, x₂ ≤ 6, x₁, x₂ ≥ 0. ## Short answer Slack variables (s₁,s₂,s₃) turn ≤ inequalities into equations, measuring unused resource. Initial BFS: s₁=18, s₂=4, s₃=6. Pivot x₂ in then x₁; final table x₁=2, x₂=6, max Z=36. ## Answer **Slack variables.** A slack variable is a non-negative variable added to the left side of a ≤ constraint to convert it into an equation. It measures the unused amount of the resource. For example, 3x₁ + 2x₂ ≤ 18 becomes 3x₁ + 2x₂ + s₁ = 18 with s₁ ≥ 0. Slack variables have zero coefficient in the objective function and provide the starting basic feasible solution (all original variables = 0). **Step 1: Standard form.** Every constraint is of the ≤ type, so add one slack variable to each (s₁, s₂, s₃ ≥ 0) to turn it into an equation. Slack variables have zero cost. ```text Maximize Z = 3x₁ + 5x₂ subject to 3x₁ + 2x₂ + s₁ = 18 x₁ + s₂ = 4 x₂ + s₃ = 6 ``` **Step 2: Initial basic feasible solution.** Put x₁, x₂ = 0, so the slack variables are the basic variables: s₁ = 18, s₂ = 4, s₃ = 6. 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 3 5 0 0 0 Basis CB x₁ x₂ s₁ s₂ s₃ b Ratio ------------------------------------ s₁ 0 3 2 1 0 0 18 9 s₂ 0 1 0 0 1 0 4 - s₃ 0 0 1 0 0 1 6 6 ------------------------------------ Zj 0 0 0 0 0 0 Cj-Zj 3 5 0 0 0 Entering: x₂ Leaving: s₃ Pivot = 1 ``` **Table 2** ```text Cj 3 5 0 0 0 Basis CB x₁ x₂ s₁ s₂ s₃ b Ratio ------------------------------------ s₁ 0 3 0 1 0 -2 6 2 s₂ 0 1 0 0 1 0 4 4 x₂ 5 0 1 0 0 1 6 - ------------------------------------ Zj 0 5 0 0 5 30 Cj-Zj 3 0 0 0 -5 Entering: x₁ Leaving: s₁ Pivot = 3 ``` **Table 3** ```text Cj 3 5 0 0 0 Basis CB x₁ x₂ s₁ s₂ s₃ b ---------------------------------- x₁ 3 1 0 1/3 0 -2/3 2 s₂ 0 0 0 -1/3 1 2/3 2 x₂ 5 0 1 0 0 1 6 ---------------------------------- Zj 3 5 1 0 3 36 Cj-Zj 0 0 -1 0 -3 ``` **Step 4: Optimality test.** In the last table every Cj − Zj is ≤ 0, so the solution is optimal. **Answer:** x₁ = 2, x₂ = 6, and Maximum Z = 36. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does a slack variable represent? A. Excess above a >= requirement B. Unused amount of the resource in a <= constraint C. An extra decision variable with profit D. Penalty cost Answer: B. It is non-negative and has zero objective coefficient. (Key point: Slack variable) 2. Which variable enters in Table 1 and which leaves? A. x2 enters, s1 leaves B. x1 enters, s2 leaves C. x2 enters, s3 leaves D. x1 enters, s1 leaves Answer: C. Largest Cj - Zj is 5 and ratio 6 is smallest (18/2 = 9, 6/1 = 6). (Key point: First pivot) 3. Why is the ratio shown as '-' for s2 in Table 1? A. The x2 column entry is 0, so no ratio exists B. s2 is artificial C. The ratio is negative infinity D. b is zero Answer: A. Division by 0 is not allowed. (Key point: Ratio rule) 4. What is the optimum? A. x1 = 0, x2 = 6, Z = 30 B. x1 = 2, x2 = 6, Z = 36 C. x1 = 4, x2 = 3, Z = 27 D. x1 = 2, x2 = 6, Z = 30 Answer: B. 3(2) + 5(6) = 36. 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