--- title: "Graphical method max:z=2x₁+3x₂" subject: "Mathematics III" chapter: 5 key: "semester_3/mathematics_iii:graphical-method-max-z-2x-3x" source: https://www.bctengineer.com/q/semester_3/mathematics_iii/graphical-method-max-z-2x-3x site: BCT Engineering --- # Graphical method max:z=2x₁+3x₂ Mathematics III · Chapter 5: Linear Programming - Asked in: 2069 Asar, 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-method-max-z-2x-3x - Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-5 ## Question Maximize z = 2x₁ + 3x₂ subject to constraints x₁ − x₂ ≤ 2, x₁ + x₂ ≥ 4 and x₁, x₂ ≥ 0 graphically. ## Short answer The feasible region is unbounded. Along the edge x₁ − x₂ = 2, Z = 5t − 6 grows without limit as t → ∞, so Z has no finite maximum. ## 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 ≤ 2 | x₁ − x₂ = 2 | (2, 0) | (0, −2) | towards the origin | | (2) x₁ + x₂ ≥ 4 | x₁ + x₂ = 4 | (4, 0) | (0, 4) | 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 9|............................................ |............................................ |............................................ |............................................ |.......................................... |........................................ |..................................... |.................................. |................................ |B............................ | ....................... | ................. | ............ | ...... | A | 0| +-------------------------------------------- 0 9 ``` **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 = 2x₁ + 3x₂ | |---|---|---|---| | A | (3, 1) | (1), (2) | 9 | | B | (0, 4) | (2), x₁ = 0 | 12 | **Step 4: Is the region bounded?** The corner points are A(3, 1) and B(0, 4) (the origin and the axis points are cut off by x₁ + x₂ ≥ 4 and x₁ − x₂ ≤ 2). Beyond them the feasible region extends to infinity: no constraint stops x₁ and x₂ from increasing together. For instance, along the line x₁ − x₂ = 2 the points (t, t − 2), t ≥ 3, are all feasible, and there Z = 2t + 3(t − 2) = 5t − 6, which grows without limit as t → ∞. Moving the line Z = constant further from the origin never leaves the region. **Answer:** The objective function is unbounded above: Z has no finite maximum. (The corner values are Z = 9 at A(3, 1) and Z = 12 at B(0, 4), but Z can be increased indefinitely along the unbounded edge.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How do you decide which side of the line x1 + x2 = 4 is allowed for x1 + x2 >= 4? A. Test the origin: 0 >= 4 is false, so the allowed side is away from the origin B. Always shade towards the origin C. Test the origin: 0 >= 4 is true, so shade towards the origin D. Shade whichever side has the larger area Answer: A. The origin does not satisfy 0 + 0 >= 4, so the feasible side is the one away from the origin. (Key point: Choosing the allowed side) 2. Which are the corner points of the feasible region? A. (0, 0) and (4, 0) B. (2, 0) and (0, 4) C. (4, 0) and (0, 2) D. A(3, 1) and B(0, 4) Answer: D. A(3, 1) is where x1 - x2 = 2 meets x1 + x2 = 4, and B(0, 4) is where x1 + x2 = 4 meets x1 = 0. (Key point: Corner points) 3. Along x1 - x2 = 2, the points (t, t-2) with t >= 3 are feasible and give Z = 5t - 6. What is the conclusion? A. Max Z = 12 at B(0, 4) B. Max Z = 9 at A(3, 1) C. The problem has no feasible solution D. Z is unbounded above, so there is no finite maximum Answer: D. Z = 5t - 6 grows without limit as t increases, so the region is unbounded in the direction of improvement and no finite maximum exists. 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