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Question
Solve the following LPP:
Maximize z = 4x1 + 3x2 subject to
3x1 + x2 ≤ 15, 3x1 + 4x2 ≤ 24, x1 ≥ 0, x2 ≥ 0.
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Solution
We first draw the lines AB and CD whose equations are 3x1 + x2 = 15 and 3x1 + 4x2 = 24 respectively.
| Line | Equation | Points on the X-axis | Points on the Y-axis | Sign | Region |
| AB | 3x1 + x2 = 15 | A(5, 0) | B(0,15) | ≤ | origin side of the line AB |
| CD | 3x1 + 4x2 = 24 | C(8, 0) | D(0, 6) | ≤ | origin side of the line CD |
The feasible region is OAPDO which is shaded in the graph.
The Vertices of the feasible region are O(0, 0), A(5, 0), P and D(0, 6).
P is the point of intersection of lines.
3x1 + 4x2 = 24 ....(1)
and 3x1 + x2 = 15 ....(2)
On subtracting, we get
3x2 = 9 ∴ x2 = 3
Substituting x2 = 3 in (2), we get
3x1 + 3 = 15
∴ 3x1 = 12
∴ x1 = 4
∴ P is (4, 3)
The values of objective function z = 4x1 + 3x2 at these vertices are
z(O) = 4(0) + 3(0) = 0 + 0 = 0
z(a) = 4(5) + 3(0) = 20 + 0 = 20
z(P) = 4(4) + 3(3) = 16 + 9 = 25
z(D) = 4(0) + 3(6) = 0 + 18 = 18
∴ z has maximum value 25 when x = 4 and y = 3.
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