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प्रश्न
Solve the following linear programming problem graphically:
Minimize: Z = 5x + 10y
Subject to constraints:
x + 2y ≤ 120, x + y ≥ 60, x – 2y ≥ 0, x ≥ 0, y ≥ 0.
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उत्तर
Here, objective function
Min. Z = 5x + 10y
Subject to constraints
x + 2y ≤ 120,
x + y ≥ 60
x – 2y ≥ 0
x ≥ 0, y ≥ 0
Changing inequations to equations, we get
x + 2y = 120 ...(i)
| x | 0 | 120 |
| y | 60 | 0 |
x + y = 60 ...(ii)
| x | 0 | 60 |
| y | 60 | 0 |
x – 2y = 0 ...(iii)
| x | 0 | 120 |
| y | 0 | 60 |
On solving equations (i) and (ii), we get point of intersection E(0, 60).
On solving equations (ii) and (iii), we get point of intersection C(40, 20).
On solving equations (i) and (iii), we get point of intersection B(60, 30).
| Corner Points | Z = 5x + 10y |
| A(120, 0) | Z = 5 × 120 + 10 × 0 = 600 |
| B(60, 30) | Z = 5 × 60 + 10 × 30 = 300 + 300 = 600 |
| C(40, 20) | Z = 5 × 40 + 10 × 20 = 200 + 200 = 400 |
| D(60, 0) | Z = 5 × 60 + 0 × 0 = 300 `rightarrow` Min. |
Hence, min. value Z is 300, when x = 60 and y = 0.
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