--- title: "Simplex method max:Z=10x₁+x₂+2x₃" subject: "Mathematics III" chapter: 5 key: "semester_3/mathematics_iii:simplex-method-max-z-10x-x" source: https://www.bctengineer.com/q/semester_3/mathematics_iii/simplex-method-max-z-10x-x site: BCT Engineering --- # Simplex method max:Z=10x₁+x₂+2x₃ Mathematics III · Chapter 5: Linear Programming - Asked in: 2079 Bhadra, 7 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-10x-x - Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-5 ## Question Solve the linear programming problem by the simplex method: Maximize Z = 10x₁ + x₂ + 2x₃ subject to x₁ + x₂ − 2x₃ ≤ 10, 4x₁ + x₂ + x₃ ≤ 20 and x₁, x₂, x₃ ≥ 0. ## Short answer Standard simplex method with slack variables leads to an optimal solution: x1 = 5, x2 = 0, x3 = 0, maximizing Z to 50. ## 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 = 10x₁ + x₂ + 2x₃ subject to x₁ + x₂ − 2x₃ + s₁ = 10 4x₁ + x₂ + x₃ + s₂ = 20 ``` **Step 2: Initial basic feasible solution.** Put x₁, x₂, x₃ = 0, so the slack variables are the basic variables: s₁ = 10, s₂ = 20. 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 10 1 2 0 0 Basis CB x₁ x₂ x₃ s₁ s₂ b Ratio ------------------------------------ s₁ 0 1 1 -2 1 0 10 10 s₂ 0 4 1 1 0 1 20 5 ------------------------------------ Zj 0 0 0 0 0 0 Cj-Zj 10 1 2 0 0 Entering: x₁ Leaving: s₂ Pivot = 4 ``` **Table 2** ```text Cj 10 1 2 0 0 Basis CB x₁ x₂ x₃ s₁ s₂ b ------------------------------------ s₁ 0 0 3/4 -9/4 1 -1/4 5 x₁ 10 1 1/4 1/4 0 1/4 5 ------------------------------------ Zj 10 5/2 5/2 0 5/2 50 Cj-Zj 0 -3/2 -1/2 0 -5/2 ``` **Step 4: Optimality test.** In the last table every Cj − Zj is ≤ 0, so the solution is optimal. **Answer:** x₁ = 5, x₂ = 0, x₃ = 0, and Maximum Z = 50. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How are the <= constraints converted to equations in the simplex method? A. Add a slack variable with zero cost to each B. Subtract a surplus variable from each C. Add an artificial variable with cost -M D. Multiply each by -1 Answer: A. Each <= row gets a slack variable with zero cost, and these give the starting basis. (Key point: Standard form) 2. In Table 1 the Cj - Zj values are 10, 1 and 2. Which variable enters? A. x2 B. x3 C. x1 D. s1 Answer: C. The entering variable is the one with the largest positive Cj - Zj, which is x1 (10). (Key point: Entering variable) 3. For entering x1, the ratios are 10/1 = 10 and 20/4 = 5. Which variable leaves? A. s1 B. s2 C. x1 D. x2 Answer: B. The smallest non-negative ratio is 5, in the s2 row, so s2 leaves with pivot 4. (Key point: Leaving variable) 4. How do we know Table 2 is optimal? A. All b values are positive B. All Cj - Zj <= 0 C. Slack variables are all zero D. The ratio column is empty Answer: B. For maximisation, optimality holds when every Cj - Zj <= 0, so no variable can improve Z. Result: x1 = 5, Z = 50. 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