--- title: "Graphical max/min min:z=9x+40y" subject: "Mathematics III" chapter: 5 key: "semester_3/mathematics_iii:graphical-max-min-min-z-9x" source: https://www.bctengineer.com/q/semester_3/mathematics_iii/graphical-max-min-min-z-9x site: BCT Engineering --- # Graphical max/min min:z=9x+40y Mathematics III · Chapter 5: Linear Programming - Asked in: 2071 Chaitra, 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-9x - Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-5 ## Question Graphically maximize and minimize z = 9x + 40y subject to the constraints y − x ≥ 1, y − x ≤ 3, 2 ≤ x ≤ 5. ## Short answer The feasible polygon has corners (2,3), (5,6), (5,8), (2,5). Evaluating z = 9x + 40y at each yields maximum 365 at (5,8) and minimum 138 at (2,3). ## 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) −x + y ≥ 1 | −x + y = 1 | (−1, 0) | (0, 1) | away from the origin | | (2) −x + y ≤ 3 | −x + y = 3 | (−3, 0) | (0, 3) | towards the origin | | (3) x ≥ 2 | x = 2 | (2, 0) | none (parallel to y-axis) | away from the origin | | (4) x ≤ 5 | x = 5 | (5, 0) | none (parallel to y-axis) | towards 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 9| | | .C | ..... | ......... | .............B | ............... | D............... | ........... | ....... | ... | A | | | | 0| +-------------------------------------------- 0 6 ``` **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 = 9x + 40y | |---|---|---|---| | A | (2, 3) | (1), (3) | 138 | | B | (5, 6) | (1), (4) | 285 | | C | (5, 8) | (2), (4) | 365 | | D | (2, 5) | (2), (3) | 218 | **Step 4: Conclusion.** The largest value is 365 at C(5, 8) and the smallest value is 138 at A(2, 3). **Answer:** Maximum Z = 365 at x = 5, y = 8; Minimum Z = 138 at x = 2, y = 3. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the feasible region bounded by? A. The axes and y - x = 3 B. A triangle formed by x = 2, x = 5 and y = 0 C. A parallelogram between y - x = 1 and y - x = 3 and between x = 2 and x = 5 D. An unbounded region above y - x = 3 Answer: C. Two pairs of parallel lines bound the region, giving a parallelogram. (Key point: Shape of the region) 2. How is D = (2, 5) obtained? A. From y - x = 3 and x = 2 B. From y - x = 1 and x = 2 C. From y - x = 3 and x = 5 D. From y - x = 1 and x = 5 Answer: A. Putting x = 2 in y - x = 3 gives y = 5. (Key point: Corner D) 3. Where is Z = 9x + 40y maximum? A. A(2, 3), Z = 138 B. C(5, 8), Z = 365 C. B(5, 6), Z = 285 D. D(2, 5), Z = 218 Answer: B. The corner values are 138, 285, 365, 218, so the maximum is 365 at C(5, 8), where y - x = 3 meets x = 5. (Key point: Maximum) 4. Where is the minimum, and why? A. At C, as it has the largest y B. At D(2, 5), Z = 218 C. At B(5, 6), Z = 285 D. At A(2, 3), Z = 138, the smallest corner value Answer: D. A(2, 3) comes from y - x = 1 and x = 2 and gives 18 + 120 = 138, the least of the corners. 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