Question

Find the feasible solution of the following inequation:

3x + 4y ≥ 12, 4x + 7y ≤ 28, y ≥ 1, x ≥ 0.

Answer 1

First we draw the lines AB, CD and EF whose equations are 3x + 4y = 12 and  4x + 7y = 28 and y= 1 respectively.

Line	Equation	Points on the X-axis	Points on the Y-axis	Sign	Region
AB	3x + 4y = 12	A (4,0)	B (0,3)	≥	non-origin side of line AB
CB	4x + 7y = 28	C (7,0)	D(0,4)	≤	origin side of line CD
EF	y = 1	-	F(0,1)	≥	non-origin side of line EF

The feasible solution is PQDB which is shaded in the graph.