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प्रश्न
Minimize: Z = 6x + 4y
Subject to the conditions:
3x + 2y ≥ 12,
x + y ≥ 5,
0 ≤ x ≤ 4,
0 ≤ y ≤ 4
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उत्तर
Graph the Feasible Region
(i) 3x + 2y ≥ 12
3x + 2y = 12 ⇒ intercepts: `{:{(x=0=> y=6),(y=0=>x=4):}`
(ii) x + y ≥ 5
x+y=5⇒intercepts: `{:{(x=0=>y=5),(y=0=>y=5):}`
(iii) 0 ≤ x ≤ 4, 0 ≤ y ≤ 4
This defines a square bounded by (0,0), (4,0), (4,4), and (0,4).
Find Corner Points of the Feasible Region
-
x + y ≥ 5
-
3x + 2y ≥ 12
1) 3x + 2y = 12 and x + y = 5
From (2): y = 5 − x
3x + 2(5 − x) = 12 ⇒ 3x + 10 − 2x = 12 ⇒ x = 2, y = 3
Point A = (2, 3)
2) 3x + 2y = 12 and y = 4
3x + 2(4) = 12 ⇒ 3x = 4 ⇒ x = `4/3, y =4`
Point B = `(4/3, 4)`
3) x + y = 5 and y = 4
x + 4 = 5 ⇒ x = 1, y = 4
Point C = (1, 4)
Points where boundary lines meet square limits:
(4, 0): Check constraints
3(4) + 2(0) = 12
4 + 0 = 4 → Doesn't satisfy x + y ≥ 5
(4, 1):
3(4) + 2(1) = 14
4 + 1 = 5
Point D = (4, 1)
Evaluate Z = 6x + 4y at all feasible corner points
| Point | Z = 6x + 4y |
| A = (2, 3) | 6(2) + 4(3) = 12 + 12 = 24 |
| B = (4/3, 4) | 6(4/3) + 4(4) = 8 + 16 = 24 |
| C = (1, 4) | 6(1) + 4(4) = 6 + 16 = 22 |
| D = (4, 1) | 6(4) + 4(1) = 24 + 4 = 28 |
The minimum value of Z is 22, which occurs at point C = (1, 4).
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