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Question
Solve the following LPP by graphical method:
Maximize z = 11x + 8y, subject to x ≤ 4, y ≤ 6, x + y ≤ 6, x ≥ 0, y ≥ 0
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Solution
First we draw the lines AB, CD and ED whose equations are x = 4, y = 6 and x + y = 6 respectively.
| Line | Equation | Points on the X-axis | Points on the Y-axis | Sign | Region |
| AB | x = 4 | A(4, 0) | - | ≤ | origin side of the line AB |
| CD | y = 6 | - | D(0, 6) | ≤ | origin side of the line CD |
| EF | x + y = 6 | E(6, 0) | D(0, 6) | ≤ | origin side of the line ED |
The feasible region is the shaded portion OAPDO in the graph.
The vertices of the feasible region are O (0, 0), A (4, 0), P and D (0, 6)
P is point of intersection of lines x + y = 6 and x = 4.
Substituting x = 4 in x + y = 6, we get
4 + y = 6 ∴ y = 2 ∴ P is (4, 2)
∴ the corner points of feasible region are O (0, 0), A (4, 0), P (4, 2) and D (0, 6).
The values of the objective function z = 11x + 8y at these vertices are
z(O) = 11(0) + 8(0) = 0 + 0 = 0
z(a) = 11(4) + 8(0) = 44 + 0 = 44
z(P) = 11(4) + 8(2) = 44 + 16 = 60
z(D) = 11(0) + 8(6) = 48
∴ z has maximum value 60, when x = 4 and y = 2.
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