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Question
Solve the following LPP:
Maximize z = 5x1 + 6x2 subject to 2x1 + 3x2 ≤ 18, 2x1 + x2 ≤ 12, x1 ≥ 0, x2 ≥ 0.
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Solution
First we draw the lines AB and CD whose equations are 2x1 + 3x2 = 18 and 2x1 + x2 = 12 respectively.
| Line | Equation | Points on the X-axis | Points on the Y-axis | Sign | Region |
| AB | 2x1 + 3x2 = 18 | A(9, 0) | B(0, 6) | ≤ | origin side of line AB |
| CD | 2x1 + x2 = 12 | C(6, 0) | O(0, 12) | ≤ | 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 (6, 0), P and B (0, 6).
P is the point of intersection of the lines
2x1 + 3x2 = 18 ....(1)
and 2x1 + x2 = 12 ....(2)
On subtracting, we get
2x2 = 6
∴ x2 = 3
Substituting x2 = 3 in (2), we get
2x1 + 3 = 12
∴ x1 = 9
∴ P is `(9/2, 3)`
The values of objective function z = 5x1 + 6x2 at these vertices are
z(O) = 5(0) + 6(0) = 0 + 0 = 0
z(C) = 5(6) + 6(0) = 30 + 0 = 30
z(P) = `5(9/2) + 6(3) = 45/2 + 18 = (45 + 36)/2 = 81/2 = 40.5`
z(B) = 5(0) + 6(3) = 0 + 18 = 18
Maximum value of z is 40.5 when x1 = `9/2` y1 = 3.
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