Find the feasible solution of the following inequation:
3x + 4y ≥ 12, 4x + 7y ≤ 28, y ≥ 1, x ≥ 0.
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.