**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.

#### Solution

To draw the feasible region, construct table as follows:

Inequality | x + 2y ≥ 3 | x + 4y ≥ 4 | 3x + y ≥ 3 |

Corresponding equation (of line) | x + 2y = 3 | x + 4y = 4 | 3x + y = 3 |

Intersection of line with X-axis | (3, 0) | (4, 0) | (1, 0) |

Intersection of line with Y-axis | `(0, 3/2)` | (0, 1) | (0, 3) |

Region | Non-origin side | Non-origin side | Non-origin side |

Shaded portion XABCDY is the feasible region,

whose vertices are A(4, 0), B, C and D(0, 3).

B is the point of intersection of the lines x + 4y = 4 and x + 2y = 3

Solving the above equations, we get

x = 2, y = `(1)/(2)` ∴ B ≡ `(2, 1/2)`

C is the point of intersection of the lines x + 2y = 3 and 3x + y = 3.

Solving the above equations, we get

x = `(3)/(5), y = (6)/(5)` ∴ C ≡ `(3/5, 6/5)`

Here, the objective function is Z = 6x + 2y

Z at A(4, 0) = 6(4) + 2(0) = 24

Z at B`(2, 1/2) = 6(2) + 2(1/2)` = 12 + 1 = 13

Z at C`(3/5, 6/5) = 6(3/5) + 2(6/5) = (18)/(5)) + (12)/(5)` = 6

∴ Z at D(0, 3) = 6(0) + 2(3) = 6

∴ Z has minimum value 6 at C`(3/5, 6/5)` and D(0, 3).

∴ Z is minimum when, x = `(3/5), y = (6/5)`, Z = 6 and x = 0, y = 3, Z = 6.