#### Question

Minimize : Z = 6x + 4y

Subject to the conditions:

3x + 2y ≥ 12,

x + y ≥ 5,

0 ≤ x ≤ 4,

0 ≤ y ≤ 4

#### 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 ABCD is the feasible region, whose vertices are A, B, C and D.

A is the point of intersection of the lines x = 4 and x + y = 5.

Putting x = 4 in x + y = 5, we get

4 + y = 5

y = 1

A = (4, 1)

B is the point of intersection of the lines y = 4 and x = 4.

B = (4, 4)

C is the point of intersection of the lines y = 4 and 3x + 2y = 12.

Putting y = 4 in 3x + 2y = 12, we get

3x + 2(4) = 12

x = 4/3

C≡(4/3,3 )

D is the point of intersection of the lines x + y = 5 and 3x + 2y = 12.

Solving the above equations, we get

D≡(2,3)

A≡(4,1),B≡(4,4),C≡(4/3,3 ) and D≡(2,3)

Here, the objective function is

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) = 24

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

Thus, Z is minimized at every point along the line segment CD and its minimum value is 24.

Therefore, Z has infinite number of optimal solutions.