--- title: "Simplex method max:Z=4x+3y" subject: "Mathematics III" chapter: 5 key: "semester_3/mathematics_iii:simplex-method-max-z-4x-3y" source: https://www.bctengineer.com/q/semester_3/mathematics_iii/simplex-method-max-z-4x-3y site: BCT Engineering --- # Simplex method max:Z=4x+3y Mathematics III · Chapter 5: Linear Programming - Asked in: 2074 Chaitra, 10 marks - Video: [Simplex Method Class 12 | Simplex Method Maximization | Class 12 Linear Programming Problem](https://www.youtube.com/watch?v=bjZI60DGl3o) — MindYourChoices - 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-z-4x-3y - 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 Z = 4x + 3y subject to 2x + 3y ≤ 6, −x + 2y ≤ 3, 2y ≤ 5, 2x + y ≤ 4, x, y ≥ 0. ## Short answer After two simplex iterations the optimal solution is x = 3/2, y = 1, yielding maximum Z = 9. ## Answer **Step 1: Standard form.** Every constraint is of the ≤ type, so add one slack variable to each (s₁, s₂, s₃, s₄ ≥ 0) to turn it into an equation. Slack variables have zero cost. ```text Maximize Z = 4x + 3y subject to 2x + 3y + s₁ = 6 −x + 2y + s₂ = 3 2y + s₃ = 5 2x + y + s₄ = 4 ``` **Step 2: Initial basic feasible solution.** Put x, y = 0, so the slack variables are the basic variables: s₁ = 6, s₂ = 3, s₃ = 5, s₄ = 4. Then Z = 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 4 3 0 0 0 0 Basis CB x y s₁ s₂ s₃ s₄ b Ratio ------------------------------------- s₁ 0 2 3 1 0 0 0 6 3 s₂ 0 -1 2 0 1 0 0 3 - s₃ 0 0 2 0 0 1 0 5 - s₄ 0 2 1 0 0 0 1 4 2 ------------------------------------- Zj 0 0 0 0 0 0 0 Cj-Zj 4 3 0 0 0 0 Entering: x Leaving: s₄ Pivot = 2 ``` **Table 2** ```text Cj 4 3 0 0 0 0 Basis CB x y s₁ s₂ s₃ s₄ b Ratio --------------------------------------- s₁ 0 0 2 1 0 0 -1 2 1 s₂ 0 0 5/2 0 1 0 1/2 5 2 s₃ 0 0 2 0 0 1 0 5 5/2 x 4 1 1/2 0 0 0 1/2 2 4 --------------------------------------- Zj 4 2 0 0 0 2 8 Cj-Zj 0 1 0 0 0 -2 Entering: y Leaving: s₁ Pivot = 2 ``` **Table 3** ```text Cj 4 3 0 0 0 0 Basis CB x y s₁ s₂ s₃ s₄ b ------------------------------------ y 3 0 1 1/2 0 0 -1/2 1 s₂ 0 0 0 -5/4 1 0 7/4 5/2 s₃ 0 0 0 -1 0 1 1 3 x 4 1 0 -1/4 0 0 3/4 3/2 ------------------------------------ Zj 4 3 1/2 0 0 3/2 9 Cj-Zj 0 0 -1/2 0 0 -3/2 ``` **Step 4: Optimality test.** In the last table every Cj − Zj is ≤ 0, so the solution is optimal. **Answer:** x = 3/2, y = 1, and Maximum Z = 9. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In Table 1, x enters (column 2, -1, 0, 2 with b = 6, 3, 5, 4). Which leaves? A. s1 B. s4 C. s2 D. s3 Answer: B. Only positive entries give ratios: 6/2 = 3 and 4/2 = 2; the minimum is 2, so s4 leaves. (Key point: Leaving variable) 2. Why do the s2 and s3 rows show '-' in the ratio column of Table 1? A. Their column entries are negative or zero B. Their b values are zero C. They are artificial variables D. They are already optimal Answer: A. Ratios are formed only for positive entries; -1 and 0 cannot limit the entering variable. (Key point: Dash in ratios) 3. In Table 2 y enters. The ratios are 1 (s1), 2 (s2), 5/2 (s3), 4 (x). Which leaves? A. s2 B. x C. s1 D. s3 Answer: C. The smallest ratio is 1, in the s1 row, so s1 leaves with pivot 2. (Key point: Second iteration) 4. What is the optimal solution? A. x = 2, y = 0, Z = 8 B. x = 3/2, y = 1, Z = 9 C. x = 0, y = 2, Z = 6 D. x = 1, y = 3/2, Z = 8.5 Answer: B. All Cj - Zj <= 0 in Table 3; check 2(3/2) + 1 = 4 and Z = 6 + 3 = 9. 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