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Question
Solve the following LPP:
Minimize z = 4x + 2y
Subject to 3x + y ≥ 27, x + y ≥ 21, x + 2y ≥ 30, x ≥ 0, y ≥ 0
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Solution
We first draw the lines AB, CD and EF whose equations are 3x + y = 27, x + y = 21, x + 2y = 30 respectively.
| Line | Equation | Points on the X-axis |
Points on the Y-axis |
Sign | Region |
| AB | 3x + y = 27 | A(9, 0) | B(0, 27) | ≥ | non-origin side of line AB |
| CD | x + y = 21 | C(21, 0) | D(0, 21) | ≥ | non-origin side of line CD |
| EF | x + 2y = 30 | E(30, 0) | F(0, 15) | ≥ | non-origin side of line EF |
The feasible region is XEPQBY which is shaded in the graph.
The vertices of the feasible region are E(30, 0), P, Q and B(0, 27).
P is the point of intersection of the lines
x + 2y = 30 ....(1)
and x + y = 21 ....(2)
On subtracting, we get
y = 9
Substituting y = 9 in (2), we get
x + 9 = 21
∴ x = 12
∴ P is (12, 9)
Q is the point of intersection of the lines
x + y = 21 ....(2)
and 3x + y = 27 ....(3)
On subtracting, we get
2x = 6
∴ x = 3
Substituting x = 3 in (2), we get
3 + y = 21
∴ y = 18
∴ Q is (3, 18)
The values of the objective function z = 4x + 2y at these vertices are
z(E) = 4(30) + 2(0) = 120 + 0 = 120
z(P) = 4(12) + 2(9) = 48 + 18 = 66
z(Q) = 4(3) + 2(18) = 12 + 36 = 48
z(B) = 4(0) + 2(27) = 0 + 54 = 54
∴ z has minimum value 48, when x = 3 and y = 18.
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