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 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.