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Question
Solve the following L.P.P. by graphical method:
Minimize: z = 8x + 10y
Subject to: 2x + y ≥ 7, 2x + 3y ≥ 15, y ≥ 2, x ≥ 0, y ≥ 0.
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Solution
First we draw the lines AB, CD and EF whose equations are 2x + y = 7, 2x + 3y = 15 and y = 2 respectively.
| Line | Equation | Points on the X-axis | Points on the Y-axis | Sign | Region |
| AB | 2x + y = 7 | A(3.5, 0) | B(0, 7) | ≥ | non-origin side of line AB |
| CD | 2x + 3y = 15 | C(7.5, 0) | D(0, 5) | ≥ | non-origin side of line CD |
| EF | y = 2 | – | F(0, 2) | ≥ | non-origin side of line EF |
The feasible region is EPQBY which is shaded in the graph. The vertices of the feasible region are P, Q and B(0, 7). P is the point of intersection of the lines 2x + 3y = 15 and y = 2.
Substituting y = 2 in 2x + 3y = 15, we get
2x + 3(2) = 15
∴ 2x = 9
∴ x = 4.5
∴ P = (4.5, 2)
Q is the point of intersection of the lines
2x + 3y = 15 .....(1)
and 2x + y = 7 .....(2)
On subtracting, we get
2y = 8
∴ y = 4
∴ from (2), 2x + 4 = 7
∴ 2x = 3
∴ x = 1.5
∴ Q = (1.5, 4)
The values of the objective function z = 8x + 10y at these vertices are
z(P) = 8(4.5) + 10(2)
= 36 + 20
= 56
z(Q) = 8(1.5) + 10(4)
= 12 + 40
= 52
z(B) = 8(0) + 10(7)
= 0 + 70
= 70
∴ z has minimum value 52, when x = 1.5 and y = 4.
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