--- title: "Simplex method max:P=x₁+x₂" subject: "Mathematics III" chapter: 5 key: "semester_3/mathematics_iii:simplex-method-max-p-x-x" source: https://www.bctengineer.com/q/semester_3/mathematics_iii/simplex-method-max-p-x-x site: BCT Engineering --- # Simplex method max:P=x₁+x₂ Mathematics III · Chapter 5: Linear Programming - Asked in: 2078 Bhadra, 7 marks - Video: [LPP WITH SIMPLEX METHOD||MAXIMIZATION MODEL WITH 3 CONSTRAINTS](https://www.youtube.com/watch?v=bmUXoXDpMrc) — SwarupaLR Vlogs - 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-x-x - Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-5 ## Question Solve the following LPP by the simplex method: Maximize P = x₁ + x₂ subject to 2x₁ + x₂ ≤ 16, x₁ ≤ 6, x₂ ≤ 10, x₁, x₂ ≥ 0. ## Short answer Using the simplex method by adding slack variables, the optimal solution is found after four iterations. The maximum value is P = 13, achieved at x₁ = 3 and x₂ = 10. ## Answer **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 P = x₁ + x₂ subject to 2x₁ + x₂ + s₁ = 16 x₁ + s₂ = 6 x₂ + s₃ = 10 ``` **Step 2: Initial basic feasible solution.** Put x₁, x₂ = 0, so the slack variables are the basic variables: s₁ = 16, s₂ = 6, s₃ = 10. 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 1 1 0 0 0 Basis CB x₁ x₂ s₁ s₂ s₃ b Ratio ------------------------------------ s₁ 0 2 1 1 0 0 16 8 s₂ 0 1 0 0 1 0 6 6 s₃ 0 0 1 0 0 1 10 - ------------------------------------ Zj 0 0 0 0 0 0 Cj-Zj 1 1 0 0 0 Entering: x₁ Leaving: s₂ Pivot = 1 ``` **Table 2** ```text Cj 1 1 0 0 0 Basis CB x₁ x₂ s₁ s₂ s₃ b Ratio ------------------------------------ s₁ 0 0 1 1 -2 0 4 4 x₁ 1 1 0 0 1 0 6 - s₃ 0 0 1 0 0 1 10 10 ------------------------------------ Zj 1 0 0 1 0 6 Cj-Zj 0 1 0 -1 0 Entering: x₂ Leaving: s₁ Pivot = 1 ``` **Table 3** ```text Cj 1 1 0 0 0 Basis CB x₁ x₂ s₁ s₂ s₃ b Ratio ------------------------------------ x₂ 1 0 1 1 -2 0 4 - x₁ 1 1 0 0 1 0 6 6 s₃ 0 0 0 -1 2 1 6 3 ------------------------------------ Zj 1 1 1 -1 0 10 Cj-Zj 0 0 -1 1 0 Entering: s₂ Leaving: s₃ Pivot = 2 ``` **Table 4** ```text Cj 1 1 0 0 0 Basis CB x₁ x₂ s₁ s₂ s₃ b ---------------------------------- x₂ 1 0 1 0 0 1 10 x₁ 1 1 0 1/2 0 -1/2 3 s₂ 0 0 0 -1/2 1 1/2 3 ---------------------------------- Zj 1 1 1/2 0 1/2 13 Cj-Zj 0 0 -1/2 0 -1/2 ``` **Step 4: Optimality test.** In the last table every Cj − Zj is ≤ 0, so the solution is optimal. **Answer:** x₁ = 3, x₂ = 10, and Maximum P = 13. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the initial basic feasible solution? A. x1 = 6, x2 = 10, P = 16 B. s1 = 16, s2 = 6, s3 = 10 with x1 = x2 = 0, P = 0 C. s1 = 0, s2 = 0, s3 = 0 D. x1 = 8, x2 = 0, P = 8 Answer: B. Setting x1 = x2 = 0 makes the slacks the basic variables equal to the right-hand sides, and P = 0. (Key point: Initial solution) 2. In Table 1, x1 enters (ties with x2 on Cj - Zj = 1). Ratios: 16/2 = 8, 6/1 = 6, and none for s3. Which variable leaves? A. s1 B. s2 C. s3 D. x2 Answer: B. The minimum ratio is 6, in the s2 row, so s2 leaves. (Key point: Leaving variable) 3. Why is no ratio computed for the s3 row in Table 1? A. The entering column entry is 0, so that row never limits the increase B. The b value is zero C. s3 is not a basic variable D. The ratio is always negative for the last row Answer: A. Ratios are taken only where the column entry is positive; a 0 entry places no limit on the entering variable. (Key point: Dash in ratio column) 4. What is the optimal solution? A. x1 = 6, x2 = 4, P = 10 B. x1 = 3, x2 = 10, P = 13 C. x1 = 0, x2 = 10, P = 10 D. x1 = 6, x2 = 10, P = 16 Answer: B. Check: 2(3) + 10 = 16 <= 16, 3 <= 6, 10 <= 10, and P = 13; all Cj - Zj <= 0. 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