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- 2069 Asar · 5 marks
Graphical method max:z=2x₁+3x₂
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.
Maximize z = 2x₁ + 3x₂ subject to constraints x₁ − x₂ ≤ 2, x₁ + x₂ ≥ 4 and x₁, x₂ ≥ 0 graphically.
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, ….
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.)
✓ Answer checked by Claude Opus 5.5 (high effort)
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