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प्रश्न
Find the feasible solution of linear inequation 2x + 3y ≤ 12, 2x + y ≤ 8, x ≥ 0, y ≥ 0 by graphically
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उत्तर
To find the feasible solution, construct the table as follows:
| Inequation | Equation | Double intercept form |
Points (x, y) |
Region |
| 2x + 3y ≤ 12 | 2x + 3y = 12 | `x/6 + y/4` = 1 | A(6, 0) B(0, 4) |
2(0) + 3(0) ≤ 12 ∴ 0 ≤ 12 ∴ origin side |
| 2x + y ≤ 8 | 2x + y = 8 | `x/4 + y/8` = 1 | C(4, 0) D(0, 8) |
2(0) + 0 ≤ 8 ∴ 0 ≤ 8 ∴ origin side |
| x ≥ 0 | x = 0 | − | − | R.H.S. of Y-axis |
| y ≥ 0 | y = 0 | − | − | Above X-axis |
The shaded portion represents the graphical solution.
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Figure
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