Question

Minimize z = 2x + 4y is subjected to 2x + y ≥ 3, x + 2y ≥ 6, x ≥ 0, y ≥ 0 show that the minimum value of z occurs at more than two points

Answer 1

To draw the feasible region, construct table as follows:

Inequality	2x + y ≥ 3	x + 2y ≥ 6
Corresponding equation (of line)	2x + y = 3	x + 2y = 6
Intersection of line with X-axis	`(3/2, 0)`	(6, 0)
Intersection of line with Y-axis	(0, 3)	(0, 3)
Region	Non-origin side	Non-origin side

x ≥ 0, y ≥ 0 represent 1^st quadrant.

Shaded portion XABY is the feasible region, whose vertices are A(6, 0) and B(0, 3).

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

∴ Z at A(6, 0) = 2(6) + 4(0) = 12

Z at B(0, 3) = 2(0) + 4(3) = 12

∴ Z is minimum at every point along the line segment AB and its minimum value is 12.

Therefore, Z has minimum value at more than two points.