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Minimize : Z = 6x + 4y

Subject to the conditions:

3x + 2y ≥ 12,

x + y ≥ 5,

0 ≤ x ≤ 4,

0 ≤ y ≤ 4

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#### Solution

To draw feasible region, construct table as follows:

Inequality | 3x+2y ≥12 | x+y ≥ 5 | x ≤ 4 | y ≤ 4 |

Corresponding equation (of line) |
3x + 2y = 12 | x+y=5 | x=4 | y=4 |

Intersection of line with X-axis |
(4, 0) | (5, 0) | (4, 0) | - |

Intersection of line with Y-axis |
(0, 6) | (0, 5) | - | (0, 4) |

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

Shaded portion is the feasible region,

Where B(4, 4), A(4, 1), C`(4/3 , 4)`

For D : 3x + 2y = 12

x + y = 5

Multiply equation (ii) by ‘2’ and subtract it from (i),

3x + 2y = 12

2x + 2y = 10

`(- - -)/(x = 2)`

∴ y = 3 and hence D (2,3)

Z = 6x + 4y

Z at A (4,1) = 6(4) + 4(1) = 28

Z at B(4,4) = 6(4) + 4(4) = 40

Z at C `(4/3 , 4) = 6 (4/3) + 4(4) = 28`

Z at D (2,3) = 6(2) + 4(3) = 24

Thus, Z is minimized at D (2,3) and its minimum value is 24.

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