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Question
Solve the following LPP by graphical method:
Maximize z = 4x + 6y, subject to 3x + 2y ≤ 12, x + y ≥ 4, x, y ≥ 0.
Solution
First we draw the lines AB, AD whose equations are 3x + 2y = 12 and x + y = 4 respectively.
| Line | Equation | Points on the X-axis | Points on the Y-axis | Sign | Region |
| AB | 3x + 2y = 12 | A(4, 0) | B(0, 6) | ≤ | origin side of the line AB |
| AC | x + y = 4 | A(4, 0) | C(0, 4) | ≥ | non-origin side of line AC |
The feasible region is the Δ ABC which is shaded in the graph.
The vertices of the feasible region (i.e. corner points) are A(4, 0), B (0, 6) and C (0, 4).
The values of the objective function z = 4x + 6y at these vertices are
z(a) = 4(4) + 6(0) = 16 + 0 = 16
z(B) = 4(0) + 6(6) = 0 + 36 = 36
z(C) = 4(0) + 6(4) = 0 + 24 = 24
∴ z has maximum value 36, when x = 0 and y = 6.
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