--- title: "Graphical method max:Z=7x₁+10x₂" subject: "Mathematics III" chapter: 5 key: "semester_3/mathematics_iii:graphical-method-max-z-7x-10x" source: https://www.bctengineer.com/q/semester_3/mathematics_iii/graphical-method-max-z-7x-10x site: BCT Engineering --- # Graphical method max:Z=7x₁+10x₂ Mathematics III · Chapter 5: Linear Programming - Asked in: 2075 Asoj, 5 marks; 2074 Asoj, 5 marks; 2073 Chaitra, 5 marks; 2072 Chaitra, 5 marks (asked 4 times) - Video: [Graphical Method for LP Problem | Graphical + Corner Point Method | Operations Research | MBA](https://www.youtube.com/watch?v=fWHzKpqZctE) — KN Prasad Academy (NITian) - 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-z-7x-10x - Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-5 ## Question Graphically maximize Z = 7x₁ + 10x₂ subject to the constraints 3x₁ + x₂ ≤ 9, x₁ + 2x₂ ≤ 8, x₁, x₂ ≥ 0. ## Short answer Plot constraints 3x1+x2≤9, x1+2x2≤8 with non-negativity. Corner points: (0,0), (3,0), (2,3), (0,4). Z=7x1+10x2 maximum 44 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) 3x₁ + x₂ ≤ 9 | 3x₁ + x₂ = 9 | (3, 0) | (0, 9) | towards the origin | | (2) x₁ + 2x₂ ≤ 8 | x₁ + 2x₂ = 8 | (8, 0) | (0, 4) | 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 5| | | |D |...... |............. |................... C |....................... |........................ |......................... |.......................... |........................... |............................ |............................. |............................... |................................ 0|A...............................B +-------------------------------------------- 0 4 ``` **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 = 7x₁ + 10x₂ | |---|---|---|---| | A | (0, 0) | x₁ = 0, x₂ = 0 | 0 | | B | (3, 0) | (1), x₂ = 0 | 21 | | C | (2, 3) | (1), (2) | 44 | | D | (0, 4) | (2), x₁ = 0 | 40 | **Step 4: Conclusion.** The largest value of Z is 44, at C(2, 3). **Answer:** Maximum Z = 44 at x₁ = 2, x₂ = 3. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the allowed side of 3x1 + x2 <= 9 decided? A. Always above the line B. Test a point on the line C. Always away from the origin D. Test the origin Answer: D. The origin satisfies 0 <= 9, so the region is toward it. (Key point: Allowed side) 2. C is the intersection of which lines, and its coordinates? A. 3x1 + x2 = 9 and x1 = 0: (0, 9) B. 3x1 + x2 = 9 and x1 + 2x2 = 8: (2, 3) C. x1 + 2x2 = 8 and x2 = 0: (8, 0) D. x1 + 2x2 = 8 and x1 = 0: (2, 3) Answer: B. Solving: 3(2)+3 = 9 and 2 + 6 = 8. (Key point: Corner point C) 3. Why are only corner points checked? A. Only corners are feasible B. A linear function over a convex polygon attains its optimum at a corner C. Z is constant inside D. Edges have no feasible points Answer: B. This is the fundamental theorem of LP. (Key point: Optimum principle) 4. What is the maximum Z? A. 40 at (0, 4) B. 21 at (3, 0) C. 80 at (0, 8) D. 44 at (2, 3) Answer: D. 7(2) + 10(3) = 44 beats 40, 21 and 0. 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