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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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| Point | Coordinates | Z = 9x + 13y | Values | Remark |
| O | (0, 0) | 9(0) + 13(0) | 0 | |
| A | (0, 6) | 9(0) + 13(6) | ______ | |
| P | ( ___,___ ) | 9( ___ ) + 13( ___ ) | ______ | ______ |
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