--- title: "Graphical method: max and min" subject: "Mathematics III" chapter: 5 key: "semester_3/mathematics_iii:graphical-method-max-and-min-two" source: https://www.bctengineer.com/q/semester_3/mathematics_iii/graphical-method-max-and-min-two site: BCT Engineering --- # Graphical method: max and min Mathematics III · Chapter 5: Linear Programming - Asked in: 2075 Chaitra, 5 marks - Video: [Linear Programming 1: Maximization -Extreme/Corner Points (LP)](https://www.youtube.com/watch?v=0TD9EQcheZM) — Joshua Emmanuel - 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-and-min-two - Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-5 ## Question Find the maximum and minimum values of the function Z = 50x₁ + 80x₂ subject to x₁ + 2x₂ ≤ 32, 3x₁ + 4x₂ ≤ 84, x₁, x₂ ≥ 0 by the graphical method. ## Short answer Plot the constraints, identify corner points of the feasible region, and evaluate Z = 50x₁ + 80x₂. Maximum Z = 1480 at (20, 6); 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) x₁ + 2x₂ ≤ 32 | x₁ + 2x₂ = 32 | (32, 0) | (0, 16) | towards the origin | | (2) 3x₁ + 4x₂ ≤ 84 | 3x₁ + 4x₂ = 84 | (28, 0) | (0, 21) | 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 18| | |D |.... |....... |.......... |............. |................ |................... |...................... |......................... |...........................C |.............................. |................................ |.................................. |.................................... 0|A.....................................B +-------------------------------------------- 0 32 ``` **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 = 50x₁ + 80x₂ | |---|---|---|---| | A | (0, 0) | x₁ = 0, x₂ = 0 | 0 | | B | (28, 0) | (2), x₂ = 0 | 1400 | | C | (20, 6) | (1), (2) | 1480 | | D | (0, 16) | (1), x₁ = 0 | 1280 | **Step 4: Conclusion.** The largest value is 1480 at C(20, 6) and the smallest value is 0 at A(0, 0). **Answer:** Maximum Z = 1480 at x₁ = 20, x₂ = 6; 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. How is the corner point C = (20, 6) found? A. Intersection of x1 + 2x2 = 32 and 3x1 + 4x2 = 84 B. Intersection of x1 = 0 and 3x1 + 4x2 = 84 C. Midpoint of the two intercepts D. Intersection of x2 = 0 and x1 + 2x2 = 32 Answer: A. Solving the two constraint lines together gives x1 = 20, x2 = 6. (Key point: Corner point C) 2. Where is Z = 50x1 + 80x2 minimum? A. At D(0, 16) with Z = 1280 B. At B(28, 0) with Z = 1400 C. At A(0, 0) with Z = 0 D. At C(20, 6) with Z = 1480 Answer: C. The origin is a feasible corner and gives Z = 0, the smallest corner value. (Key point: Minimum) 3. Which corner gives the maximum? A. C(20, 6), Z = 1480 B. B(28, 0), Z = 1400 C. D(0, 16), Z = 1280 D. A(0, 0), Z = 0 Answer: A. Comparing 0, 1400, 1480, 1280, the largest is 1480 at C. 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