--- title: "Graphical method max:Z=5x₁+3x₂" subject: "Mathematics III" chapter: 5 key: "semester_3/mathematics_iii:graphical-method-max-z-5x-3x" source: https://www.bctengineer.com/q/semester_3/mathematics_iii/graphical-method-max-z-5x-3x site: BCT Engineering --- # Graphical method max:Z=5x₁+3x₂ Mathematics III · Chapter 5: Linear Programming - Asked in: 2074 Chaitra, 5 marks; 2073 Shrawan, 5 marks; 2071 Shrawan, 5 marks (asked 3 times) - Video: [[#1] LPP - Graphical method [ Maximization with 2 constraints ] solved problem :-by kauserwise](https://www.youtube.com/watch?v=8IRrgDoV8Eo) — Kauser Wise - 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-5x-3x - Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-5 ## Question Graphically maximize Z = 5x₁ + 3x₂ subject to the constraints x₁ + 2x₂ ≤ 50, 2x₁ + x₂ ≤ 40, x₁, x₂ ≥ 0. ## Short answer Plot constraints x1+2x2≤50, 2x1+x2≤40, x1,x2≥0. Corner points: (0,0), (20,0), (10,20), (0,25). Max Z=5x1+3x2 is 110 at (10,20). ## 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) x₁ + 2x₂ ≤ 50 | x₁ + 2x₂ = 50 | (50, 0) | (0, 25) | towards the origin | | (2) 2x₁ + x₂ ≤ 40 | 2x₁ + x₂ = 40 | (20, 0) | (0, 40) | 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 29| | |D |...... |............. |...................C |..................... |....................... |........................ |.......................... |............................ |............................. |............................... |................................. |................................... |.................................... 0|A....................................B +-------------------------------------------- 0 23 ``` **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 = 5x₁ + 3x₂ | |---|---|---|---| | A | (0, 0) | x₁ = 0, x₂ = 0 | 0 | | B | (20, 0) | (2), x₂ = 0 | 100 | | C | (10, 20) | (1), (2) | 110 | | D | (0, 25) | (1), x₁ = 0 | 75 | **Step 4: Conclusion.** The largest value of Z is 110, at C(10, 20). **Answer:** Maximum Z = 110 at x₁ = 10, x₂ = 20. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What are the intercepts of 2x1 + x2 = 40? A. (50, 0) and (0, 25) B. (20, 0) and (0, 20) C. (20, 0) and (0, 40) D. (40, 0) and (0, 20) Answer: C. Set each variable to zero in turn. (Key point: Intercepts) 2. Where do the two constraint lines meet? A. (0, 25) B. (20, 0) C. (20, 10) D. (10, 20) Answer: D. Solve x1 + 2x2 = 50 and 2x1 + x2 = 40: x1 = 10, x2 = 20. (Key point: Intersection) 3. Which corner gives the maximum Z = 5x1 + 3x2? A. B(20, 0) with Z = 100 B. D(0, 25) with Z = 75 C. C(10, 20) with Z = 110 D. A(0, 0) Answer: C. 5(10) + 3(20) = 110 is the largest. 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