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Question
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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Solution
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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