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प्रश्न
Maximize: z = 3x + 5y Subject to
x +4y ≤ 24 3x + y ≤ 21
x + y ≤ 9 x ≥ 0 , y ≥0
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उत्तर
| Inequation | Point on x-axis | Point on y-axis | Feasible Region |
| x +4y ≤ 24 | (24,0) | (0,6) | Origin side |
| 3x +y ≤ 21 | (7,0) | (0,21) | Origin side |
| x + y ≤9 | (9,0) | (0,9) | Origin side |
From the figure common feasible region is ABCDEA
E is the point of intersection of x + y = 9 and x + 4y =24
Solving them we get E(4,5)
D is the point of intersection of x + y = 9 and 3x +y = 21
Solving them we get D(6,3)
| End Point | Value of z = 3x +5y |
| A(0,6) | 0 +30= 30 |
| B(0,0) | 0 + 0 = 0 |
| C(7,0) | 21 + 0 = 21 |
| D(6,3) | 18 +1 5 = 33 |
| E(4,5) | 12 +25 = 37 |
∴ z is maximum 37 at the point (4, 5)
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