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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.
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.
संबंधित प्रश्न
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| Machine | Products | ||
| A | B | C | |
| M1 M2 |
4 | 3 | 5 |
| 2 | 2 | 4 | |
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Solve the LPP graphically:
Minimize Z = 4x + 5y
Subject to the constraints 5x + y ≥ 10, x + y ≥ 6, x + 4y ≥ 12, x, y ≥ 0
Solution: Convert the constraints into equations and find the intercept made by each one of it.
| 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.
∴ Solution of the inequations is away from origin.
The feasible region is unbounded area which is satisfied by all constraints.
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 ___
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Maximised value of z in z = 3x + 4y, subject to constraints : x + y ≤ 4, x ≥ 0. y ≥ 0
If z = 200x + 500y .....(i)
Subject to the constraints:
x + 2y ≥ 10 .......(ii)
3x + 4y ≤ 24 ......(iii)
x, 0, y ≥ 0 ......(iv)
At which point minimum value of Z is attained.
