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प्रश्न
Solve the following L.P.P. by graphical method:
Maximize: Z = 10x + 25y
subject to 0 ≤ x ≤ 3,
0 ≤ y ≤ 3,
x + y ≤ 5.
Also find the maximum value of z.
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उत्तर
To draw the feasible region, construct table as follows:
| Inequality | x ≤ 3 | y ≤ 3 | x + y ≤ 5 |
| Corresponding equation (of line) | x = 3 | y = 3 | x + y = 5 |
| Intersection of line with X-axis | (3, 0) | – | (5, 0) |
| Intersection of line with Y-axis | – | (0, 3) | (0, 5) |
| Region | Origin side | Origin side | Origin side |
Shaded portion OABCD is the feasible region,
whose vertices are O(0, 0), A(3, 0), B, C and D(0, 3)
B is the point of intersection of the lines x = 3 and x + y = 5.
Substituting x = 3 in x + y = 5, we get y = 2
∴ B ≡ (3, 2)
C is the point of intersection of the lines y = 3 and x + y = 5.
Substituting y = 3 in x + y = 5, we get
x = 2
∴ C ≡ (2, 3)
Here, the objective function is Z = 10x + 25y
∴ Z at O(0, 0) = 10(0) + 25(0) = 0
Z at A(3, 0) = 10(3) + 25(0) = 30
Z at B(3, 2) = 10(3) + 25(2) = 30 + 50 = 80
Z at C(2, 3) = 10(2) + 25(3) = 20 + 75 = 95
Z at D(0, 3) = 10(0) + 25(3) = 75
∴ Z has a maximum value of 95 at C(2, 3).
∴ Z is maximum when x = 2 and y = 3.
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| 5x + y ≥ 10 | 5x + y = 10 | ( ___, 0) | (0, 10) | Away from origin |
| x + y ≥ 6 | x + y = 6 | (6, 0) | (0, ___ ) | Away from origin |
| x + 4y ≥ 12 | x + 4y = 12 | (12, 0) | (0, 3) | Away from origin |
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| A | (12, 0) | 4(12) + 5(0) | 48 | |
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| C | ( ___, ___ ) | 4( ___) + 5(___ ) | ______ | |
| D | (0, 10) | 4(0) + 5(10) | 50 |
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