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प्रश्न
A wholesale dealer deals in two kinds of mixtures A and B of nuts. Each kg of mixture A contains 60 grams of almonds, 30 grams of cashew and 30 grams of hazel nuts. Each kg of mixture B contains 30 grams of almonds, 60 grams of cashew and 180 grams of hazel nuts. A dealer is contemplating to use mixtures A and B to make a bag which will contain at least 240 grams of almonds, 300 grams of cashew and 540 grams of hazel nuts. Mixture A costs ₹ 8 and B costs ₹ 12 per kg. How many kgs of each mixture should he use to minimize the cost of the kgs
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उत्तर
Let the dealer use x kg of mixture A and y kg of mixture B.
Since x and y cannot be negative, x ≥ 0, y ≥ 0
Mixture A costs ₹ 8 and Mixture B costs ₹ 12 per kg.
∴ Total cost = Z = 8x + 12y
We construct a table with the constraints of Almond, Cashew and Hazelnut as follows:
| Mixture A | Mixture B | Least value | |
| Almond | 60 | 30 | 240 |
| Cashew | 30 | 60 | 300 |
| Hazelnut | 30 | 180 | 540 |
From the table, the constraints are
60x + 30y ≥ 240
30x + 60y ≥ 300
30x + 180y ≥ 540
∴ Given problem can be formulated as follows:
Minimize Z = 8x + 12y
Subject to 60x + 30y ≥ 240
30x + 60y ≥ 300
30x + 180y ≥ 540, x ≥ 0, y ≥ 0
To draw the feasible region, construct table as follows:
| Inequality | 60x +30y ≥ 240 | 30x+ 60y ≥ 300 | 30x+ 180y ≥ 540 |
| Corresponding equation (of line) | 60x+ 30y = 240 | 30x + 60y = 300 | 30x+ 180y = 540 |
| Intersection of line with X-axis | (4, 0) | (10, 0) | (18, 0) |
| Intersection of line with Y-axis | (0, 8) | (0, 5) | (0, 3) |
| Region | Non-origin side | Non-origin side | Non-origin side |
Shaded portion XABCDY is the feasible region, whose vertices are A(18, 0), B, C and D(0, 8).
B is the point of intersection of the lines 30x + 180y = 540 and 30x + 60y = 300.
Solving the above equations, we get
x = 6, y = 2
∴ B ≡ (6, 2)
C is the point of intersection of the lines 60x + 30y = 240 and 30x + 60y = 300.
Solving the above equations, we get
x = 2, y = 4
∴ C ≡ (2, 4)
Here, the objective function is
Z = 8x + 12y
∴ Z at A(18, 0) = 8(18) + 12(0)
= 144
Z at B(6, 2) = 8(6) + 12(2)
= 48 + 24
= 72
Z at C(2, 4) = 8(2) + 12(4)
= 16 + 48
= 64
Z at D(0, 8) = 8(0) + 12(8)
= 96
∴ Z has minimum value 64 at x = 2 and y = 4.
∴ 2 kgs of mixture A and 4 kgs of mixture B should be used to minimize the cost of the kgs.
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| A | (12, 0) | 4(12) + 5(0) | 48 | |
| B | ( ___, ___ ) | 4( ___) + 5(___ ) | ______ | ______ |
| C | ( ___, ___ ) | 4( ___) + 5(___ ) | ______ | |
| D | (0, 10) | 4(0) + 5(10) | 50 |
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