--- title: "Simplex method max:P=30x₁+x₂" subject: "Mathematics III" chapter: 5 key: "semester_3/mathematics_iii:simplex-method-max-p-30x-x" source: https://www.bctengineer.com/q/semester_3/mathematics_iii/simplex-method-max-p-30x-x site: BCT Engineering --- # Simplex method max:P=30x₁+x₂ Mathematics III · Chapter 5: Linear Programming - Asked in: 2076 Chaitra, 7 marks - 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-30x-x - Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-5 ## Question Use the simplex method to solve the following LPP: Maximize P = 30x₁ + x₂ subject to 2x₁ + x₂ ≤ 10, x₁ + 3x₂ ≤ 10, x₁, x₂ ≥ 0. ## Short answer Adding slack variables and applying simplex iterations yields an optimal solution in two tables. Maximum P = 150 at x₁ = 5, x₂ = 0. ## 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 P = 30x₁ + x₂ subject to 2x₁ + x₂ + 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 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 30 1 0 0 Basis CB x₁ x₂ s₁ s₂ b Ratio --------------------------------- s₁ 0 2 1 1 0 10 5 s₂ 0 1 3 0 1 10 10 --------------------------------- Zj 0 0 0 0 0 Cj-Zj 30 1 0 0 Entering: x₁ Leaving: s₁ Pivot = 2 ``` **Table 2** ```text Cj 30 1 0 0 Basis CB x₁ x₂ s₁ s₂ b ------------------------------- x₁ 30 1 1/2 1/2 0 5 s₂ 0 0 5/2 -1/2 1 5 ------------------------------- Zj 30 15 15 0 150 Cj-Zj 0 -14 -15 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 P = 150. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In Table 1 the Cj - Zj values are 30 for x1 and 1 for x2. Which enters and why? A. x2, because it has the smaller coefficient B. x1, because it has the largest positive Cj - Zj C. s1, because it is a slack D. Neither; the table is already optimal Answer: B. The rule is to pick the variable with the largest positive Cj - Zj, here x1 = 30. (Key point: Entering variable) 2. For entering x1, the ratios are 10/2 = 5 and 10/1 = 10. Which row leaves? A. s1 (ratio 5) B. s2 (ratio 10) C. Both leave D. Neither Answer: A. The smallest non-negative ratio is 5, so s1 leaves with pivot 2. (Key point: Leaving variable) 3. In Table 2, Cj - Zj = -14 for x2 and -15 for s1. What does this show? A. The solution can be improved by x2 B. The solution is infeasible C. The problem is unbounded D. The solution is optimal as all values are <= 0 Answer: D. Every Cj - Zj <= 0, so no variable can raise P. The answer is x1 = 5, x2 = 0, P = 150. 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