Question

Minimize Z = x + 4y subject to constraints

x + 3y ≥ 3, 2x + y ≥ 2, x ≥ 0, y ≥ 0

Answer 1

To draw the feasible region, construct table as follows:

Inequality	x + 3y ≥ 3	2x + y ≥ 2
Corresponding equation (of line)	x + 3y = 3	2x + y = 2
Intersection of line with X-axis	(3, 0)	(1, 0)
Intersection of line with Y-axis	(0, 1)	(0, 2)
Region	Non-origin side	Non-origin side

Shaded portion XABCY is the feasible region, whose vertices are A(3, 0), B and C(0, 2).

B is the point of intersection of the lines 2x + y = 2 and x + 3y = 3.

∴ B ≡ `(3/5, 4/5)`

Here, the objective function is

Z = x + 4y

∴ Z at A(3, 0) = 3 + 4(0)

= 3

Z at B`(3/5, 4/5) = 3/5 + 4(4/5)`

= `19/5`

= 3.8

Z at C(0, 2) = 0 + 4(2)

= 8

∴ Z has minimum value 3 at x = 3 and y = 0.