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प्रश्न
Solve the following LPP by graphical method:
Maximize: z = 3x + 5y
Subject to: x + 4y ≤ 24
3x + y ≤ 21
x + y ≤ 9
x ≥ 0, y ≥ 0
Also find the maximum value of z.
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उत्तर
To draw the feasible region, construct table as follows:
| Inequality | x + 4y ≤ 24 | 3x + y ≤ 21 | x + y ≤ 9 |
| Corresponding equation (of line) | x + 4y = 24 | 3x + y = 21 | x + y = 9 |
| Intersection of line with X-axis | (24, 0) | (7, 0) | (9, 0) |
| Intersection of line with Y-axis | (0, 6) | (0, 21) | (0, 9) |
| Region | Origin side | Origin side | Origin side |
Shaded portion OABCD is the feasible region.
Whose vertices are O(0, 0), A(7, 0), B, C and (0, 6)
B is the point of intersection of the lines 3x + y = 21 and x + y = 9.
Solving the above equations, we get x = 6, y = 3
∴ B ≡ (6, 3)
C is the point of intersection of the lines x + 4y = 24 and x + y = 9.
Solving the above equations, we get
x = 4, y = 5
∴ C ≡ (4, 5)
Here, the objective function is Z = 3x + 5y.
Z at O(0, 0) = 3(0) + 5(0) = 0
Z at A(7, 0) = 3(7) + 5(0) = 21
Z at B(6, 3) = 3(6) + 5(3) = 18 + 15 = 33
Z at C(4, 5) = 3(4) + 5(5) = 12 + 25 = 37
Z at D(0, 6) = 3(0) + 5(6) = 30
∴ Z has a maximum value of 37 at C(4, 5).
∴ Z is maximum when x = 4, y = 5.
संबंधित प्रश्न
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|
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