--- title: "Graphical method max:Z=x₁+1.5x₂" subject: "Mathematics III" chapter: 5 key: "semester_3/mathematics_iii:graphical-method-max-z-x-1" source: https://www.bctengineer.com/q/semester_3/mathematics_iii/graphical-method-max-z-x-1 site: BCT Engineering --- # Graphical method max:Z=x₁+1.5x₂ Mathematics III · Chapter 5: Linear Programming - Asked in: 2068 Baishakh, 5 marks - Video: [Graphical Method for LP Problem | Graphical + Corner Point Method | Operations Research | MBA](https://www.youtube.com/watch?v=fWHzKpqZctE) — KN Prasad Academy (NITian) - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/mathematics_iii/graphical-method-max-z-x-1 - Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-5 ## Question Maximize Z = x₁ + 1.5x₂ subject to the constraints 2x₁ + 2x₂ ≤ 160, x₁ + 2x₂ ≤ 120, 4x₁ + 2x₂ ≤ 280, x₁ ≥ 0 and x₂ ≥ 0 graphically. ## Short answer Plot the three constraints and find corner points: (0,0), (70,0), (60,20), (40,40), (0,60). Evaluating Z = x₁ + 1.5x₂ gives maximum 100 at x₁ = 40, x₂ = 40. ## Answer **Step 1: Draw the constraint lines.** Replace each inequality by an equation, find the intercepts and decide which side of the line is allowed (test the origin). | Constraint | Line | Meets x-axis | Meets y-axis | Allowed side | |---|---|---|---|---| | (1) 2x₁ + 2x₂ ≤ 160 | 2x₁ + 2x₂ = 160 | (80, 0) | (0, 80) | towards the origin | | (2) x₁ + 2x₂ ≤ 120 | x₁ + 2x₂ = 120 | (120, 0) | (0, 60) | towards the origin | | (3) 4x₁ + 2x₂ ≤ 280 | 4x₁ + 2x₂ = 280 | (70, 0) | (0, 140) | towards the origin | Also x₁ ≥ 0 and x₂ ≥ 0, so the region lies in the first quadrant. **Step 2: Feasible region.** The region satisfying all constraints is the shaded polygon below (dots), with corner points marked A, B, C, …. ```text 68| | |E |...... |.......... |............... |.................... |......................D |.......................... |............................ |.............................. |................................ C |.................................. |................................... |.................................... |..................................... 0|A.....................................B +-------------------------------------------- 0 79 ``` **Step 3: Corner points and the objective value.** The optimum of a linear function over a convex polygon occurs at a corner point. | Point | Coordinates | Lines meeting there | Z = x₁ + 1.5x₂ | |---|---|---|---| | A | (0, 0) | x₁ = 0, x₂ = 0 | 0 | | B | (70, 0) | (3), x₂ = 0 | 70 | | C | (60, 20) | (1), (3) | 90 | | D | (40, 40) | (1), (2) | 100 | | E | (0, 60) | (2), x₁ = 0 | 90 | **Step 4: Conclusion.** The largest value of Z is 100, at D(40, 40). **Answer:** Maximum Z = 100 at x₁ = 40, x₂ = 40. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What are the intercepts of 4x1 + 2x2 = 280? A. (70, 0) and (0, 140) B. (140, 0) and (0, 70) C. (280, 0) and (0, 280) D. (80, 0) and (0, 80) Answer: A. x2 = 0 gives x1 = 280/4 = 70; x1 = 0 gives x2 = 280/2 = 140. (Key point: Intercepts) 2. Which two constraint lines meet at D, and what are its coordinates? A. Lines (1) and (3) at (60, 20) B. Lines (1) and (2) at (40, 40) C. Lines (2) and (3) at (40, 40) D. Line (2) and x1 = 0 at (0, 60) Answer: B. Solving 2x1 + 2x2 = 160 and x1 + 2x2 = 120 gives x1 = 40, x2 = 40. (Key point: Corner D from intersection) 3. Corner values of Z = x1 + 1.5x2 are 0, 70, 90 at C(60,20), 100 at D(40,40) and 90 at E(0,60). What is the answer? A. Max Z = 90 at (60, 20) B. Max Z = 100 at (40, 40) C. Max Z = 90 at (0, 60) D. Max Z = 70 at (70, 0) Answer: B. The largest corner value is 100 at D(40, 40), so that is the maximum. 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