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Question
Solve the following L.P.P. by graphical method :
Maximize: Z = 3x + 5y subject to x + 4y ≤ 24, 3x + y ≤ 21, x + y ≤ 9, x ≥ 0, y ≥ 0 also find maximum value of Z.
Solution
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 D (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 = 4y = 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 maximum value 37 at C(4, 5).
∴ Z is maximum, when x = 4, y = 5.
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