--- title: "Graphical method min:z=4x₁+3x₂+x₃" subject: "Mathematics III" chapter: 5 key: "semester_3/mathematics_iii:graphical-method-min-z-4x-3x" source: https://www.bctengineer.com/q/semester_3/mathematics_iii/graphical-method-min-z-4x-3x site: BCT Engineering --- # Graphical method min:z=4x₁+3x₂+x₃ Mathematics III · Chapter 5: Linear Programming - Asked in: 2070 Chaitra, 5 marks - 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-min-z-4x-3x - Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-5 ## Question Graphically minimize z = 4x₁ + 3x₂ + x₃ subject to x₁ + 2x₂ + 4x₃ ≥ 12, 3x₁ + 2x₂ + x₃ ≥ 8 and x₁, x₂, x₃ ≥ 0. ## Short answer The three-variable primal is solved via its dual, which has two variables and is drawn graphically. The dual maximum is W = 8 at w₁ = 0, w₂ = 1; by duality the primal optimum is x₁ = 0, x₂ = 0, x₃ = 8, with Min z = 8. ## Answer **Reading of the question.** The problem as printed has three variables (x₁, x₂, x₃), so its feasible region lies in three-dimensional space and cannot be drawn on a plane. It is solved here graphically through its dual, which has only two variables (one per constraint) and can be drawn. The solution is the one for the problem exactly as printed. **Step 1: Write the dual.** Primal: Minimize z = 4x₁ + 3x₂ + x₃ subject to x₁ + 2x₂ + 4x₃ ≥ 12, 3x₁ + 2x₂ + x₃ ≥ 8, x₁, x₂, x₃ ≥ 0. The dual is: ```text Maximize W = 12w₁ + 8w₂ subject to w₁ + 3w₂ ≤ 4 2w₁ + 2w₂ ≤ 3 4w₁ + w₂ ≤ 1 w₁, w₂ ≥ 0 ``` (Each primal variable gives one dual constraint: the coefficients of x₁ are (1, 3), of x₂ are (2, 2) and of x₃ are (4, 1), and the costs 4, 3, 1 become the right-hand sides.) **Step 2: Draw the dual 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) w₁ + 3w₂ ≤ 4 | w₁ + 3w₂ = 4 | (4, 0) | (0, 4/3) | towards the origin | | (2) 2w₁ + 2w₂ ≤ 3 | 2w₁ + 2w₂ = 3 | (3/2, 0) | (0, 3/2) | towards the origin | | (3) 4w₁ + w₂ ≤ 1 | 4w₁ + w₂ = 1 | (1/4, 0) | (0, 1) | towards the origin | Also w₁ ≥ 0 and w₂ ≥ 0, so the region lies in the first quadrant. **Step 3: Feasible region of the dual.** The region satisfying all constraints is the shaded polygon below (dots), with corner points marked A, B, C, …. ```text 2| | | | | | | | |C |.. |... |..... |...... |....... |......... |.......... 0|A..........B +-------------------------------------------- 0 1 ``` **Step 4: Corner points and the value of W.** The optimum of a linear function over a convex polygon occurs at a corner point. | Point | Coordinates | Lines meeting there | W = 12w₁ + 8w₂ | |---|---|---|---| | A | (0, 0) | w₁ = 0, w₂ = 0 | 0 | | B | (1/4, 0) | (3), w₂ = 0 | 3 | | C | (0, 1) | (3), w₁ = 0 | 8 | **Step 5: Conclusion.** The largest W is 8 at (w₁, w₂) = (0, 1). By the duality theorem, Min z = Max W = 8. **Step 6: Primal values (complementary slackness).** At (0, 1) the first two dual constraints are not tight (0 + 3 = 3 < 4 and 0 + 2 = 2 < 3), so x₁ = 0 and x₂ = 0. Since w₂ = 1 > 0, the second primal constraint is tight: 3(0) + 2(0) + x₃ = 8, so x₃ = 8. The first primal constraint is then 4(8) = 32 ≥ 12, which holds with w₁ = 0. Check: z = 4(0) + 3(0) + 1(8) = 8, and all primal constraints are satisfied. **Answer:** Minimum z = 8 at x₁ = 0, x₂ = 0, x₃ = 8. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is this three-variable problem solved graphically through its dual? A. The dual always has a smaller optimal value B. The primal has no feasible solution C. The primal has three variables and cannot be drawn on a plane, while the dual has only two variables D. Graphical methods only work for maximization problems Answer: C. A three-variable region needs 3D space; the dual has one variable per primal constraint, i.e. two, so it can be drawn. (Key point: Why use the dual) 2. What are the dual constraints for this primal (columns of x1, x2, x3 are (1,3), (2,2), (4,1); costs 4, 3, 1)? A. w1 + 3w2 <= 4, 2w1 + 2w2 <= 3, 4w1 + w2 <= 1 B. w1 + 3w2 >= 4, 2w1 + 2w2 >= 3, 4w1 + w2 >= 1 C. w1 + 2w2 + 4w3 <= 12 and 3w1 + 2w2 + w3 <= 8 D. w1 + 2w2 <= 4, 3w1 + 2w2 <= 3, 4w1 + w2 <= 1 Answer: A. Each primal variable gives one dual constraint of type <=, using its column as coefficients and its cost as the right-hand side. (Key point: Forming the dual) 3. The largest W = 12w1 + 8w2 over the dual region is 8 at (0, 1). What does this give for the primal? A. Min z = 12 B. Min z = 8, since Min z = Max W by the duality theorem C. Max z = 8 D. Min z = 0 because the origin is feasible Answer: B. By the duality theorem the optimal values of primal and dual are equal, so Min z = Max W = 8. 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