--- title: "Simplex method max:P=15x+10y" subject: "Mathematics III" chapter: 5 key: "semester_3/mathematics_iii:simplex-method-max-p-15x-10y" source: https://www.bctengineer.com/q/semester_3/mathematics_iii/simplex-method-max-p-15x-10y site: BCT Engineering --- # Simplex method max:P=15x+10y Mathematics III · Chapter 5: Linear Programming - Asked in: 2080 Baishakh, 7 marks; 2073 Shrawan, 10 marks; 2070 Asar, 7 marks; 2068 Chaitra, 7.5 marks (asked 4 times) - 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-15x-10y - Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-5 ## Question Solve the following linear programming problem by using the simplex method: Maximize P = 15x + 10y subject to 2x + y ≤ 10, x + 3y ≤ 10 and x, y ≥ 0. ## Short answer Add slack variables, apply simplex. After two iterations, optimal tableau gives x=4, y=2, max P=80. ## 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 = 15x + 10y subject to 2x + y + s₁ = 10 x + 3y + s₂ = 10 ``` **Step 2: Initial basic feasible solution.** Put x, y = 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 15 10 0 0 Basis CB x y 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 15 10 0 0 Entering: x Leaving: s₁ Pivot = 2 ``` **Table 2** ```text Cj 15 10 0 0 Basis CB x y s₁ s₂ b Ratio -------------------------------------- x 15 1 1/2 1/2 0 5 10 s₂ 0 0 5/2 -1/2 1 5 2 -------------------------------------- Zj 15 15/2 15/2 0 75 Cj-Zj 0 5/2 -15/2 0 Entering: y Leaving: s₂ Pivot = 5/2 ``` **Table 3** ```text Cj 15 10 0 0 Basis CB x y s₁ s₂ b ------------------------------- x 15 1 0 3/5 -1/5 4 y 10 0 1 -1/5 2/5 2 ------------------------------- Zj 15 10 7 1 80 Cj-Zj 0 0 -7 -1 ``` **Step 4: Optimality test.** In the last table every Cj − Zj is ≤ 0, so the solution is optimal. **Answer:** x = 4, y = 2, and Maximum P = 80. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why are slack variables added? A. To convert <= constraints into equations B. To increase profit C. To convert >= to <= D. To make x non-negative Answer: A. They take up unused resource and have zero cost. (Key point: Slack variables) 2. Which variable enters first and why? A. s1, largest b B. x, largest Cj - Zj (15) C. y, ratio is smallest D. y, largest coefficient in constraint Answer: B. The largest positive Cj - Zj is 15 for x. (Key point: Entering variable) 3. Which leaves first? A. s1, ratio 5 is the smallest B. s1, ratio 10 C. s2, ratio 10 D. s2, ratio 2 Answer: A. 10/2 = 5 vs 10/1 = 10. (Key point: Leaving variable) 4. What is the final result? A. x = 4, y = 2, P = 70 B. x = 4, y = 2, P = 80 C. x = 2, y = 4, P = 70 D. x = 5, y = 0, P = 75 Answer: B. 15(4) + 10(2) = 80 and all Cj - Zj are <= 0. 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