--- title: "Graphical method max:Z=4x₁+5x₂" subject: "Mathematics III" chapter: 5 key: "semester_3/mathematics_iii:graphical-method-max-z-4x-5x" source: https://www.bctengineer.com/q/semester_3/mathematics_iii/graphical-method-max-z-4x-5x site: BCT Engineering --- # Graphical method max:Z=4x₁+5x₂ Mathematics III · Chapter 5: Linear Programming - Asked in: 2069 Chaitra, 5 marks - Video: [Lec-1 Graphical Method In Linear Programming Problem || For Unique Solution || In Hindi](https://www.youtube.com/watch?v=ku1KSgBfzs4) — Start Practicing - 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-4x-5x - Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-5 ## Question Maximize Z = 4x₁ + 5x₂ subject to constraints 2x₁ + 5x₂ ≤ 25, 6x₁ + 5x₂ ≤ 45, x₁ ≥ 0 and x₂ ≥ 0 graphically. ## Short answer The feasible region is a quadrilateral with vertices (0,0), (7.5,0), (5,3), (0,5). The objective Z = 4x₁ + 5x₂ attains its maximum 35 at x₁ = 5, x₂ = 3. ## 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 + 5x₂ ≤ 25 | 2x₁ + 5x₂ = 25 | (25/2, 0) | (0, 5) | towards the origin | | (2) 6x₁ + 5x₂ ≤ 45 | 6x₁ + 5x₂ = 45 | (15/2, 0) | (0, 9) | 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 6| | | |D. |...... |........... |............... |.................... |........................C |.......................... |........................... |............................. |.............................. |................................ |................................. |................................... 0|A...................................B +-------------------------------------------- 0 9 ``` **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 = 4x₁ + 5x₂ | |---|---|---|---| | A | (0, 0) | x₁ = 0, x₂ = 0 | 0 | | B | (15/2, 0) | (2), x₂ = 0 | 30 | | C | (5, 3) | (1), (2) | 35 | | D | (0, 5) | (1), x₁ = 0 | 25 | **Step 4: Conclusion.** The largest value of Z is 35, at C(5, 3). **Answer:** Maximum Z = 35 at x₁ = 5, x₂ = 3. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What are the intercepts of 6x1 + 5x2 = 45? A. (6, 0) and (0, 5) B. (45, 0) and (0, 45) C. (15/2, 0) and (0, 9) D. (9, 0) and (0, 15/2) Answer: C. Setting x2 = 0 gives x1 = 45/6 = 15/2; setting x1 = 0 gives x2 = 45/5 = 9. (Key point: Intercepts of constraint lines) 2. Where do 2x1 + 5x2 = 25 and 6x1 + 5x2 = 45 intersect? A. (3, 5) B. (5, 3) C. (0, 5) D. (15/2, 0) Answer: B. Subtracting the equations gives 4x1 = 20, so x1 = 5; then 5x2 = 15, so x2 = 3. (Key point: Intersection corner point) 3. Corner values of Z = 4x1 + 5x2 are 0 at (0,0), 30 at (15/2,0), 35 at (5,3) and 25 at (0,5). What is the answer? A. Max Z = 35 at x1 = 5, x2 = 3 B. Max Z = 30 at x1 = 15/2, x2 = 0 C. Max Z = 45 at (0, 9) D. Max Z = 25 at x1 = 0, x2 = 5 Answer: A. The optimum of a linear function over a convex polygon lies at a corner; the largest corner value is 35 at C(5, 3). 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