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Graph

**Solve the following problem :**

Minimize Z = 4x + 2y Subject to 3x + y ≥ 27, x + y ≥ 21, x ≥ 0, y ≥ 0

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#### Solution

To find the graphical solution, construct the table as follows:

Inequation | equation | Double intercept form | Points (x1, x2) | Points (x1, x2) |

3x + y ≥ 27 | 3x + y = 27 | `x/(9) + y/(27)` = 1 | A (9, 0) B (0, 27) |
3(0) + 0 ≥ 27 |

x + y ≥ 21 | x + y = 21 | `x/(21) + y/(21)` = 1 | C (21, 0) D (0, 21) |
(0) + 0 ≥ 21 ∴ 0 ≥ 21 ∴ non-origin-side |

x ≥ 0 | x = 0 | – | R.H.S. of Y-axis | |

y ≥ 0 | y = 0 | above X-axis |

The shaded portion CHB is the feasible region.

Whose vertices are C(21, 0), H and B(0, 27)

H is the point of intersection of lines

3x + y = 27 …(i)

x + y = 21 …(ii)

∴ By (i) – (ii), we get

3 x + y = 27

x + y = 21

– – –

2x = 6

∴ x = `(6)/(2)` = 3

Substituting x = 3 in (ii), we get

3 + y = 21

∴ y = 18

∴ H (3, 18)

Here, the objective function is Z = 4x + 2y

Now, we will find minimum value of Z as follows:

Feasible points | The value of Z = 4x + 2y |

C (21, 0) | Z = 4(21) + 2(0) = 84 |

H (3, 18) | Z = 4(3) + 2(18) = 12 + 36 = 48 |

B (0, 27) | Z = 4(0) + 2(27) = 54 |

∴ Z has minimum value 48 at H (3, 18)

∴ Z is minimum, when x = 3, y = 18

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