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Question
Solve the following L.P.P. by graphical method:
Minimize: Z = 6x + 2y subject to x + 2y ≥ 3, x + 4y ≥ 4, 3x + y ≥ 3, x ≥ 0, y ≥ 0.
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Solution
To draw the feasible region, construct the table as follows:
| Equation of line | Intercepts | Constraint Type | Feasible region |
| x + 2y = 3 | x = 3, y = `3/2` | ≥ | Non-origin side |
| x + 4y = 4 | x = 4, y = 1 | ≥ | Non-origin side |
| 3x + y = 3 | x = 1, y = 3 | ≥ | Non-origin side |
Pis point of intersection of the lines x + 2 y = 3 and x + 4y = 4.
Solving this equation, we will get x = 2, y = `1/2`
∴ P = `(2, 1/2)`
Q is the point of intersection of the lines x + 2 y = 3 and 3x + y = 3.
Solving this equation, we will get `x = 3/5, y = 6/5`
∴ Q = `(3/5, 6/5)`
The common shaded region is feasible region with boundary points C(4, 0), P`(2, 1/2)`
Q = `(3/5, 6/5)` and F(0,3).
Substitute these points in z = 6x + 2y
| Points | Value of z = 6x + 2y |
| C(4, 0) | ZC = 6(4) + 2(0) = 24 |
| P`(2, 1/2)` | ZP = `6(2) + 2(1/2) = 13` |
| Q = `(3/5, 6/5)` | ZQ = `6(3/5) + 2(6/5) = 6` |
| F(0,3) | ZF = 6(0) + 2(3) = 6 |
From the above tabulation, Z is minimum at two points Q and F.
∴ Zmin = 6 at `"Q"(3/5, 6/5)` and F(0,3).
∴ The L.P.P. has infinitely many solutions.
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| 5x + y ≥ 10 | 5x + y = 10 | ( ___, 0) | (0, 10) | Away from origin |
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In the figure, ABCD represents
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The coordinates of C are obtained by solving equations
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Hence the optimum solution lies at the extreme points.
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| A | (12, 0) | 4(12) + 5(0) | 48 | |
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| C | ( ___, ___ ) | 4( ___) + 5(___ ) | ______ | |
| D | (0, 10) | 4(0) + 5(10) | 50 |
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