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प्रश्न
Solve the following problem :
A factory produced two types of chemicals A and B The following table gives the units of ingredients P & Q (per kg) of Chemicals A and B as well as minimum requirements of P and Q and also cost per kg. of chemicals A and B.
| Ingredients per kg. /Chemical Units | A (x) |
B (y) |
Minimum requirements in |
| P | 1 | 2 | 80 |
| Q | 3 | 1 | 75 |
| Cost (in ₹) | 4 | 6 |
Find the number of units of chemicals A and B should be produced so as to minimize the cost.
उत्तर
Let the factory produces ‘x’ units of chemical A and ‘y’ units of chemical B
∴ Total cost Z = 4x + 6y
This is the objective function to be minimized.
From the given information, the constraints are
x + 2y ≥ 80, 3x + y ≥ 75, x ≥ 0, y ≥ 0
∴ Given problem can be formulated as
Minimize Z = 4x + 6y
Subject to, x + 2y ≥ 80, 3x + y ≥ 75, x ≥ 0, y ≥ 0
To draw the feasible region, construct table as follows:
| Inequality | x + 2y ≥ 80 | 3x + y ≥ 75 |
| Corresponding equation (of line) | x + 2y = 80 | 3x + y = 75 |
| Intersection of line with X-axis | (80, 0) | (25, 0) |
| Intersection of line with Y-axis | (0, 40) | (0, 75) |
| Region | Non-origin side | Non-origin side |
Shaded portion XABCY is the feasible region,
whose vertices are A ≡ (80, 0), B and C ≡ (0, 75)
B is the point of intersection of the lines 3x + y = 75 and x + 2y = 80
Solving the above equations, we get
B ≡ (14, 33)
Here the objective function is,
Z = 4x + 6y
∴ Z at A(80, 0) = 4(80) + 6(0) = 320
Z at B(14, 33) = 4(14) + 6(33) = 254
Z at C(0, 75) = 4(0) + 6(75) = 450
∴ Z has minimum value 254 at B(14, 33)
∴ Z is mimimum, when x = 14, y = 33.
∴ Factory should produce 14 units of chemical A and 33 units of chemical B to minimize the cost to ₹ 254.
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