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Graphical method max:Z=7x₁+10x₂
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).
Graphically maximize Z = 7x₁ + 10x₂ subject to the constraints 3x₁ + x₂ ≤ 9, x₁ + 2x₂ ≤ 8, x₁, x₂ ≥ 0.
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, ….
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.
✓ Answer checked by Claude Opus 5.5 (high effort)
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