--- title: "Graphical max/min opt:Z=4x₁+5x₂" subject: "Mathematics III" chapter: 5 key: "semester_3/mathematics_iii:graphical-max-min-opt-z-4x" source: https://www.bctengineer.com/q/semester_3/mathematics_iii/graphical-max-min-opt-z-4x site: BCT Engineering --- # Graphical max/min opt:Z=4x₁+5x₂ Mathematics III · Chapter 5: Linear Programming - Asked in: 2082 Shrawan, 5 marks - Video: [Linear Programming (Optimization) 2 Examples Minimize & Maximize](https://www.youtube.com/watch?v=Y7e7DCsDUMY) — Mario's Math Tutoring - 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-opt-z-4x - Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-5 ## Question Find the maximum and minimum values of the following linear programming problem by using the graphical method: Optimize Z = 4x₁ + 5x₂ subject to x₁ + x₂ ≤ 10, −x₁ + x₂ ≥ 2, x₁, x₂ ≥ 0. ## Short answer Draw x₁+x₂≤10, -x₁+x₂≥2, x₁,x₂≥0. Corner points: (0,2) Z=10, (4,6) Z=46, (0,10) Z=50. Maximum Z=50 at x₁=0, x₂=10; minimum Z=10 at x₁=0, x₂=2. ## 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 + x₂ ≤ 10 | x₁ + x₂ = 10 | (10, 0) | (0, 10) | towards the origin | | (2) −x₁ + x₂ ≥ 2 | −x₁ + x₂ = 2 | (−2, 0) | (0, 2) | away from 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 12| | | |C.. |......... |................ |...................... |............................ |..................................B |............................ |...................... |................ |......... |A.. | | 0| +-------------------------------------------- 0 5 ``` **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, 2) | (2), x₁ = 0 | 10 | | B | (4, 6) | (1), (2) | 46 | | C | (0, 10) | (1), x₁ = 0 | 50 | **Step 4: Conclusion.** The largest value is 50 at C(0, 10) and the smallest value is 10 at A(0, 2). **Answer:** Maximum Z = 50 at x₁ = 0, x₂ = 10; Minimum Z = 10 at x₁ = 0, x₂ = 2. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For -x1 + x2 >= 2, which side is feasible? A. Away from the origin B. Left of the y-axis C. Towards the origin D. On the line only Answer: A. The origin gives 0 >= 2, which is false. (Key point: Allowed side) 2. What is the intersection of x1 + x2 = 10 and -x1 + x2 = 2? A. (0, 10) B. (4, 6) C. (6, 4) D. (2, 0) Answer: B. Add: 2x2 = 12, x2 = 6, x1 = 4. (Key point: Corner point B) 3. Maximum Z is at which corner? A. C(0, 10), Z = 50 B. B(4, 6), Z = 46 C. B(4, 6), Z = 50 D. A(0, 2), Z = 10 Answer: A. 50 > 46 > 10. (Key point: Maximum) 4. The minimum Z is? A. 10 at (0, 2) B. 0 at origin C. 46 at (4, 6) D. 50 at (0, 10) Answer: A. The origin is not feasible, so A(0, 2) gives the smallest value. 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