--- title: "Graphical max/min min:Z=5x₁+3x₂" subject: "Mathematics III" chapter: 5 key: "semester_3/mathematics_iii:graphical-max-min-min-z-5x" source: https://www.bctengineer.com/q/semester_3/mathematics_iii/graphical-max-min-min-z-5x site: BCT Engineering --- # Graphical max/min min:Z=5x₁+3x₂ Mathematics III · Chapter 5: Linear Programming - Asked in: 2072 Kartik, 5 marks - 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-max-min-min-z-5x - Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-5 ## Question Graphically maximize and minimize Z = 5x₁ + 3x₂ subject to the constraints 3x₁ + 5x₂ ≤ 15, 5x₁ + 2x₂ ≤ 10, x₁, x₂ ≥ 0. ## Short answer Plot the lines 3x1+5x2=15 and 5x1+2x2=10 to find the feasible region. Corner points are (0,0), (2,0), (20/19, 45/19), (0,3). Evaluate Z=5x1+3x2: maximum Z=235/19 at (20/19, 45/19); minimum Z=0 at (0,0). ## 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) 3x₁ + 5x₂ ≤ 15 | 3x₁ + 5x₂ = 15 | (5, 0) | (0, 3) | towards the origin | | (2) 5x₁ + 2x₂ ≤ 10 | 5x₁ + 2x₂ = 10 | (2, 0) | (0, 5) | 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 4| | | | |D |...... |............ |...............C |.................. |................... |..................... |...................... |....................... |......................... |.......................... |............................ 0|A............................B +-------------------------------------------- 0 3 ``` **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 | (2, 0) | (2), x₂ = 0 | 10 | | C | (20/19, 45/19) | (1), (2) | 235/19 | | D | (0, 3) | (1), x₁ = 0 | 9 | **Step 4: Conclusion.** The largest value is 235/19 at C(20/19, 45/19) and the smallest value is 0 at A(0, 0). **Answer:** Maximum Z = 235/19 at x₁ = 20/19, x₂ = 45/19; Minimum Z = 0 at x₁ = 0, x₂ = 0. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Solving 3x1 + 5x2 = 15 and 5x1 + 2x2 = 10 together gives which corner point C? A. (2, 0) B. (20/19, 45/19) C. (45/19, 20/19) D. (0, 3) Answer: B. Eliminating gives 19x1 = 20, so x1 = 20/19 and then x2 = 45/19. (Key point: Corner point by solving equations) 2. What is Z = 5x1 + 3x2 at C(20/19, 45/19)? A. 10 B. 235/19 C. 9 D. 65/19 Answer: B. Z = 5(20/19) + 3(45/19) = (100 + 135)/19 = 235/19. (Key point: Value at C) 3. Corner values are 0 (A), 10 (B), 235/19 (C), 9 (D). What are the maximum and minimum? A. Max 10 at B, Min 0 at A B. Max 235/19 at C, Min 9 at D C. Max 235/19 at C, Min 0 at A D. Max 235/19 at C, Min 10 at B Answer: C. Compare all corner values: the largest is 235/19 (about 12.4) at C and the smallest is 0 at the origin A. 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