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Question
Solve the following LPP:
Maximize z =60x + 50y subject to
x + 2y ≤ 40, 3x + 2y ≤ 60, x ≥ 0, y ≥ 0.
Solution
We first draw the lines AB and CD whose equations are x + 2y = 40 and 3x + 2y = 60 respectively.
| Line | Equation | Points on the X-axis | Points on the Y-axis | Sign | Region |
| AB | x + 2y = 40 | A(40,0) | B(0,20) | ≤ | origin side of line AB |
| CD | 3x + 2y = 60 | C(20,0) | D(0,30) | ≤ | origin side of line CD |
The feasible region is OCPBO which is shaded in the graph.
The vertices of the feasible region are O (0, 0), C (20, 0), P and B (0, 20).
P is the point of intersection of the lines.
3x + 2y = 60 ....(1)
and x + 2y = 40 .....(2)
On subtracting, we get
2x = 20 ∴ x = 10
Substituting x = 10 in (2), we get
10 + 2y = 40
∴ 2y = 30
∴ y = 15
∴ P is (10, 15)
The values of the objective function z = 60x + 50y at these vertices are
z(O) = 60(0) + 50(0) = 0 + 0 = 0
z(C) = 60(20) + 50(0) = 1200 + 0 = 1200
z(P) = 60(10) + 50(15) = 600 + 750 = 1350
z(B) = 60(0) + 50(20) = 0 + 1000 = 1000
∴ z has maximum value 1350 at x = 10, y = 15.
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