--- title: "Simplex method max:P=20x₂−5x₁" subject: "Mathematics III" chapter: 5 key: "semester_3/mathematics_iii:simplex-method-max-p-20x-5x" source: https://www.bctengineer.com/q/semester_3/mathematics_iii/simplex-method-max-p-20x-5x site: BCT Engineering --- # Simplex method max:P=20x₂−5x₁ Mathematics III · Chapter 5: Linear Programming - Asked in: 2071 Chaitra, 10 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-20x-5x - Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-5 ## Question Solve the following linear programming problem by the simplex method: Maximize P = 20x₂ − 5x₁ subject to 10x₂ − 2x₁ ≤ 5, 2x₁ + 5x₂ ≤ 10 and x₁, x₂ ≥ 0. ## Short answer One simplex iteration yields the optimal basic solution x₁ = 0, x₂ = 1/2, with maximum P = 10. ## 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 = −5x₁ + 20x₂ subject to −2x₁ + 10x₂ + s₁ = 5 2x₁ + 5x₂ + s₂ = 10 ``` **Step 2: Initial basic feasible solution.** Put x₁, x₂ = 0, so the slack variables are the basic variables: s₁ = 5, 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 -5 20 0 0 Basis CB x₁ x₂ s₁ s₂ b Ratio --------------------------------- s₁ 0 -2 10 1 0 5 1/2 s₂ 0 2 5 0 1 10 2 --------------------------------- Zj 0 0 0 0 0 Cj-Zj -5 20 0 0 Entering: x₂ Leaving: s₁ Pivot = 10 ``` **Table 2** ```text Cj -5 20 0 0 Basis CB x₁ x₂ s₁ s₂ b --------------------------------- x₂ 20 -1/5 1 1/10 0 1/2 s₂ 0 3 0 -1/2 1 15/2 --------------------------------- Zj -4 20 2 0 10 Cj-Zj -1 0 -2 0 ``` **Step 4: Optimality test.** In the last table every Cj − Zj is ≤ 0, so the solution is optimal. **Answer:** x₁ = 0, x₂ = 1/2, and Maximum P = 10. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is one slack variable added to each constraint in 10x2 - 2x1 <= 5 and 2x1 + 5x2 <= 10? A. To convert each <= inequality into an equation, with slack variables having zero cost in the objective B. To make the objective function larger C. To convert the problem from maximization to minimization D. To remove the non-negativity condition on x1 and x2 Answer: A. Slack variables turn <= constraints into equalities and carry zero cost, so the objective function is unchanged. (Key point: Slack variables) 2. In Table 1 (Cj - Zj = -5, 20, 0, 0), which variable enters the basis and why? A. x1, because it is the first column B. x2, because it has the largest positive Cj - Zj (20) C. s1, because its Cj - Zj is zero D. x1, because -5 is the most negative value Answer: B. For maximization the entering variable has the largest positive Cj - Zj, which is x2 with 20. (Key point: Entering variable) 3. Ratios in Table 1 are 5/10 = 1/2 for s1 and 10/5 = 2 for s2. Which variable leaves and what is the pivot? A. s2 leaves, pivot = 5 B. s2 leaves, pivot = 2 C. s1 leaves, pivot = 10 D. s1 leaves, pivot = -2 Answer: C. The smallest non-negative ratio (1/2) belongs to s1, so s1 leaves and the pivot is the x2 entry in its row, 10. (Key point: Leaving variable by ratio) 4. In Table 2 the Cj - Zj row is -1, 0, -2, 0. What does this tell us and what is the final answer? A. Not optimal; x1 must enter next B. Optimal because all Cj - Zj <= 0; x1 = 0, x2 = 1/2, Max P = 10 C. Optimal; x1 = 1/2, x2 = 0, Max P = 10 D. Optimal because Zj = 0 in the s2 column; Max P = 15/2 Answer: B. No positive Cj - Zj remains, so the table is optimal. x2 = 1/2 gives P = 20 x 1/2 = 10 with x1 = 0. (Key point: Optimality test and result) ## Close variants set in other exams - [Simplex method max:P=15x+10y](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Asimplex-method-max-p-15x-10y) — Mathematics III, chapter 5, asked 17× - [Simplex method max:Z=3x₁+5x₂](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Asimplex-method-max-z-3x-5x-two) — Mathematics III, chapter 5, asked 17× - [Simplex method max:Z=4x+3y](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Asimplex-method-max-z-4x-3y) — Mathematics III, chapter 5, asked 17× - [Big-M method max:Z=2x₁+x₂](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Abig-m-method-max-z-2x) — Mathematics III, chapter 5, asked 17× - [Simplex method max:Z=10x₁+x₂+2x₃](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Asimplex-method-max-z-10x-x) — Mathematics III, chapter 5, asked 18× - [Simplex method max:Z=15x₁+10x₂](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Asimplex-method-max-z-15x-10x) — Mathematics III, chapter 5, asked 15× - [Big-M method max:Z=4x₁+5x₂](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Abig-m-method-max-z-4x) — Mathematics III, chapter 5, asked 15× - [Big-M method max:F=2x₁+x₂](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Abig-m-method-max-f-2x) — Mathematics III, chapter 5, asked 17× - [Two-phase simplex method max:Z=3x₁−x₂](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Atwo-phase-simplex-method-max-z) — Mathematics III, chapter 5, asked 15× - [Big-M method max:P=2x₁+5x₂](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Abig-m-method-max-p-2x-two) — Mathematics III, chapter 5, asked 15× - [Big-M method max:P=2x₁+x₂+3x₃](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Abig-m-method-max-p-2x) — Mathematics III, chapter 5, asked 17× ## Related in Mathematics III - [Simplex method max:Z=3x₁+5x₂](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Asimplex-method-max-z-3x-5x) — Mathematics III, chapter 5 - [Big-M method max:P=2x₁+5x₂](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Abig-m-method-max-p-2x-four) — Mathematics III, chapter 5, asked 3× - [Simplex method max:P=−x₁+2x₂](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Asimplex-method-max-p-x-2x) — Mathematics III, chapter 5, asked 8×