--- title: "Simplex method max:P=−x₁+2x₂" subject: "Mathematics III" chapter: 5 key: "semester_3/mathematics_iii:simplex-method-max-p-x-2x" source: https://www.bctengineer.com/q/semester_3/mathematics_iii/simplex-method-max-p-x-2x site: BCT Engineering --- # Simplex method max:P=−x₁+2x₂ Mathematics III · Chapter 5: Linear Programming - Asked in: 2079 Baishakh, 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-2x - 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 + 2x₂ subject to −x₁ + x₂ ≤ 2, −x₁ + 3x₂ ≤ 12, x₁ − 4x₂ ≤ 4, x₁, x₂ ≥ 0. ## Short answer Add slack variables s1,s2,s3 to form equations, set initial basis s1=2,s2=12,s3=4. Simplex: x2 enters, s1 leaves; then x1 enters, s2 leaves. Optimal tableau yields x1=3, x2=5, with Maximum P=7. ## 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 + 2x₂ subject to −x₁ + x₂ + s₁ = 2 −x₁ + 3x₂ + s₂ = 12 x₁ − 4x₂ + s₃ = 4 ``` **Step 2: Initial basic feasible solution.** Put x₁, x₂ = 0, so the slack variables are the basic variables: s₁ = 2, s₂ = 12, s₃ = 4. 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 2 0 0 0 Basis CB x₁ x₂ s₁ s₂ s₃ b Ratio ------------------------------------ s₁ 0 -1 1 1 0 0 2 2 s₂ 0 -1 3 0 1 0 12 4 s₃ 0 1 -4 0 0 1 4 - ------------------------------------ Zj 0 0 0 0 0 0 Cj-Zj -1 2 0 0 0 Entering: x₂ Leaving: s₁ Pivot = 1 ``` **Table 2** ```text Cj -1 2 0 0 0 Basis CB x₁ x₂ s₁ s₂ s₃ b Ratio ------------------------------------ x₂ 2 -1 1 1 0 0 2 - s₂ 0 2 0 -3 1 0 6 3 s₃ 0 -3 0 4 0 1 12 - ------------------------------------ Zj -2 2 2 0 0 4 Cj-Zj 1 0 -2 0 0 Entering: x₁ Leaving: s₂ Pivot = 2 ``` **Table 3** ```text Cj -1 2 0 0 0 Basis CB x₁ x₂ s₁ s₂ s₃ b ---------------------------------- x₂ 2 0 1 -1/2 1/2 0 5 x₁ -1 1 0 -3/2 1/2 0 3 s₃ 0 0 0 -1/2 3/2 1 21 ---------------------------------- Zj -1 2 1/2 1/2 0 7 Cj-Zj 0 0 -1/2 -1/2 0 ``` **Step 4: Optimality test.** In the last table every Cj − Zj is ≤ 0, so the solution is optimal. **Answer:** x₁ = 3, x₂ = 5, and Maximum P = 7. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How do we convert x1 - 4x2 <= 4 into an equation for the simplex method? A. Subtract a surplus variable and add an artificial variable B. Add a slack variable s3: x1 - 4x2 + s3 = 4 C. Subtract a slack variable: x1 - 4x2 - s3 = 4 D. Multiply by -1 and treat it as >= Answer: B. Every <= constraint gets one slack variable (cost zero), which also forms the initial basis. (Key point: Standard form) 2. In Table 1 the x2 column is 1, 3, -4 and b is 2, 12, 4. Which row leaves and why is the s3 row skipped? A. s2 leaves; s3 ratio is 4/(-4) = -1 and is negative B. s3 leaves because it has the largest b C. s1 leaves; negative or zero column entries are not used in the ratio test, so s3 has no ratio D. s1 leaves because 2 is the smallest right-hand side Answer: C. Only positive column entries give valid ratios. s1 has ratio 2 (smallest), s2 has 4, s3 has none (-4 is not positive), so s1 leaves. (Key point: Ratio test with negative entries) 3. In Table 2, Cj - Zj = 1 for x1. What happens next? A. Stop; optimal B. x1 enters and s2 leaves (ratio 3, pivot 2) C. x1 enters and s3 leaves D. x2 leaves the basis because it has the largest value Answer: B. The positive Cj - Zj means improvement is possible; the x1 column has positive entry only in s2 (2) with ratio 6/2 = 3, so s2 leaves. (Key point: Second iteration) 4. What is the optimal solution? A. x1 = 5, x2 = 3, Max P = 1 B. x1 = 0, x2 = 2, Max P = 4 C. x1 = 3, x2 = 5, Max P = 7 D. x1 = 3, x2 = 5, Max P = 13 Answer: C. The final table gives x1 = 3, x2 = 5, so P = -3 + 2(5) = 7. 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