--- title: "Graphical method: max and min" subject: "Mathematics III" chapter: 5 key: "semester_3/mathematics_iii:graphical-method-max-and-min" source: https://www.bctengineer.com/q/semester_3/mathematics_iii/graphical-method-max-and-min site: BCT Engineering --- # Graphical method: max and min Mathematics III · Chapter 5: Linear Programming - Asked in: 2076 Asoj, 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 - Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-5 ## Question Find the maximum and minimum values of the function Z = 20x + 10y subject to x + 2y ≤ 40, 3x + y ≥ 30, 4x + 3y ≥ 60, x, y ≥ 0 by the graphical method. ## Short answer Plot constraints to find the feasible region, evaluate objective Z = 20x + 10y at corner points. Maximum Z = 800 occurs at (40, 0); Minimum Z = 240 at (6, 12). ## 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) x + 2y ≤ 40 | x + 2y = 40 | (40, 0) | (0, 20) | towards the origin | | (2) 3x + y ≥ 30 | 3x + y = 30 | (10, 0) | (0, 30) | away from the origin | | (3) 4x + 3y ≥ 60 | 4x + 3y = 60 | (15, 0) | (0, 20) | away from the origin | Also x ≥ 0 and y ≥ 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 21| | | D | . | .... | ...... | ........ | A......... | ............ | ............. | ............... | ................ | .................. | ................... | ..................... | ...................... 0| B.......................C +-------------------------------------------- 0 45 ``` **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 = 20x + 10y | |---|---|---|---| | A | (6, 12) | (2), (3) | 240 | | B | (15, 0) | (3), y = 0 | 300 | | C | (40, 0) | (1), y = 0 | 800 | | D | (4, 18) | (1), (2) | 260 | **Step 4: Conclusion.** The largest value is 800 at C(40, 0) and the smallest value is 240 at A(6, 12). **Answer:** Maximum Z = 800 at x = 40, y = 0; Minimum Z = 240 at x = 6, y = 12. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What are the axis intercepts of 3x + y = 30? A. (30, 0) and (0, 10) B. (10, 0) and (0, 30) C. (3, 0) and (0, 1) D. (15, 0) and (0, 20) Answer: B. Putting y = 0 gives x = 10 and x = 0 gives y = 30. (Key point: Drawing a constraint) 2. How do we decide which side of 3x + y >= 30 is allowed? A. Always shade towards the origin B. Always shade away from the origin C. Test the origin: 0 >= 30 is false, so shade away from it D. Shade between the intercepts Answer: C. Substituting the origin gives a false statement, so the allowed side is away from the origin. (Key point: Choosing the side) 3. Where does the maximum or minimum of Z over the feasible polygon lie? A. At a corner point B. At the centre of the region C. On any boundary point D. At the origin Answer: A. A linear function over a convex polygon attains its extremes at corner points. (Key point: Where optimum occurs) 4. What are the maximum and minimum of Z = 20x + 10y? A. Max 800 at (40, 0); min 300 at (15, 0) B. Max 365 at (5, 8); min 138 at (2, 3) C. Max 260 at (4, 18); min 240 at (6, 12) D. Max 800 at (40, 0); min 240 at (6, 12) Answer: D. Z values at corners are 240, 300, 800, 260, so the largest is 800 at C and the smallest is 240 at A. 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