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