Question

Minimize : Z = 6x + 4y

Subject to the conditions:

3x + 2y ≥ 12,

x + y ≥ 5,

0 ≤ x ≤ 4,

0 ≤ y ≤ 4

Answer 1

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.