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Question
Solve the following Linear Programming Problem graphically.
Maximise Z = 5x + 2y subject to:
x – 2y ≤ 2,
3x + 2y ≤ 12,
– 3x + 2y ≤ 3,
x ≥ 0, y ≥ 0
Solution
Given L.P.P. is
Max. Z = 5x + 2y subject to
x – 2y ≤ 2,
3x + 2y ≤ 12,
– 3x + 2y ≤ 3,
and x ≥ 0, y ≥ 0
First, we draw graphs of lines
x – 2y = 2 ...(1)
3x + 2y = 12 ...(2)
– 3x + 2y = 3 ...(3)
By (1) x – 2y = 2
| x | 2 | 0 |
| y | 0 | – 1 |
A(2, 0) and B(0, – 1)
By (2) 3x + 2y = 12
| x | 4 | 0 |
| y | 0 | 6 |
C(4, 0) and D(0, 6)
By (3) – 3x + 2y = 3
| x | – 1 | 0 |
| y | 0 | 3/2 |
`E(-1, 0), F(0, 3/2)`
Now solving (1) and (2), we get
x = `7/2` and y = `3/4`
Point of intersection `P(7/2, 3/4)`
Solving (2) and (3), we get
x = `3/2`, y = `15/4`
Point of intersection `Q(3/2, 15/4)`
x ≥ 0, y ≥ 0 shows that the shaded region lies in first quadrant.
For x – 2y ≤ 2, take origin (0, 0), then 0 – 0 ≤ 2 which is true.
∴ Shaded the region in which origin lies.
Similarly, we shade the region for various situations.
Hence OAPQFO is the required feasible region.
The corresponding Z values can be evaluated as follows:
| Vertices of shaded region |
Z = 5z + 2y |
| O(0, 0) | Z = 5 × 0 + 2 × 0 = 0 |
| A(2, 0) | Z = 10 + 0 = 10 |
| `P(7/2, 3/4)` | Z = `35/2 + 3/2` = 19 (maximum) |
| `Q(3/2, 15/4)` | Z = `15/2 + 15/2` = 15 |
| `F(0, 3/2)` | Z = 0 + 3 = 3 |
Hence, the maximum value of Z is 19 at the point `P(7/2, 3/4)`.
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