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प्रश्न
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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उत्तर
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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| Inequation | Equation | X intercept | Y intercept | Region |
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| O | (0, 0) | 9(0) + 13(0) | 0 | |
| A | (0, 6) | 9(0) + 13(6) | ______ | |
| P | ( ___,___ ) | 9( ___ ) + 13( ___ ) | ______ | ______ |
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∴ Z is maximum at __( ___, ___ ) with the value ___.
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Minimize Z = 4x + 5y
Subject to the constraints 5x + y ≥ 10, x + y ≥ 6, x + 4y ≥ 12, x, y ≥ 0
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| Inequations | Equations | X intercept | Y intercept | Region |
| 5x + y ≥ 10 | 5x + y = 10 | ( ___, 0) | (0, 10) | Away from origin |
| x + y ≥ 6 | x + y = 6 | (6, 0) | (0, ___ ) | Away from origin |
| x + 4y ≥ 12 | x + 4y = 12 | (12, 0) | (0, 3) | Away from origin |
| x, y ≥ 0 | x = 0, y = 0 | x = 0 | y = 0 | 1st quadrant |
∵ Origin has not satisfied the inequations.
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In the figure, ABCD represents
The set of the feasible solution where
A(12, 0), B( ___, ___ ), C ( ___, ___ ) and D(0, 10).
The coordinates of B are obtained by solving equations
x + 4y = 12 and x + y = 6
The coordinates of C are obtained by solving equations
5x + y = 10 and x + y = 6
Hence the optimum solution lies at the extreme points.
The optimal solution is in the following table:
| Point | Coordinates | Z = 4x + 5y | Values | Remark |
| A | (12, 0) | 4(12) + 5(0) | 48 | |
| B | ( ___, ___ ) | 4( ___) + 5(___ ) | ______ | ______ |
| C | ( ___, ___ ) | 4( ___) + 5(___ ) | ______ | |
| D | (0, 10) | 4(0) + 5(10) | 50 |
∴ Z is minimum at ___ ( ___, ___ ) with the value ___
