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Question
Solve the following problem :
A firm manufacturing two types of electrical items A and B, can make a profit of ₹ 20 per unit of A and ₹ 30 per unit of B. Both A and B make use of two essential components, a motor and a transformer. Each unit of A requires 3 motors and 2 transformers and each unit of B requires 2 motors and 4 transformers. The total supply of components per month is restricted to 210 motors and 300 transformers. How many units of A and B should be manufacture per month to maximize profit? How much is the maximum profit?
Solution
Let the firm manufacture x units of A and y units of B.
The profit is ₹ 20 per unit of A and ₹ 30 per unit of B.
∴ Total profit = ₹ (20 x + 30 y).
We construct a table with the constraints of number of motors and transformers needed.
| Electrical item\Essential component | A (x) |
B (y) |
Maximum Supply |
| Motors | 3 | 2 | 210 |
| Transformers | 2 | 4 | 300 |
From the table, the total motors required is (3x + 2y) and total motor required is (2x + 4y).
But total supply of components per month is restricted to 210 motors and 300 transformers.
∴ The constraints are 3x + 2y ≤ 210 and 2x + 4y ≤ 300.
As x, y cannot be negative, we have x ≤ 0 and y ≥ 0.
Hence the given LPP can be formulated as follows:
Maximize Z = 20x + 30y
Subject to
3x + 2y ≤ 210,
2x + 4y ≤ 300,
x ≤ 0, y ≥ 0.
For graphical solutions of the inequalities, consider lines L1 : 3x + 2y = 210 and 2x + 4y = 300
For L1 :
| x | y | (x, y) |
| 0 | 105 | (0, 105) |
| 70 | 0 | (70, 0) |
For L2 :
| x | y | (x, y) |
| 0 | 75 | (0, 75) |
| 150 | 0 | (150, 0) |
L1 passes through A (0, 105) and B (70, 0)
L2 passes through P (0, 75) and Q (150, 0)
Solving both lines, we get x = 30, y = 60
The coordinates of origin O (0, 0) satisfies both the inequalities.
∴ The required region is on origin side of both the lines L1 and L2.
As x ≥ 0, y ≥ 0; the feasible region lies in the first quadrant.
OBRP is the required feasible region.
At O (0, 0), Z = 0 + 0 = 0
At B (70, 0), Z = 20 (70) + 0 = 1400
At R (30, 60), Z = 20 (30) + 30 (60) = 2400
At P (0, 75), Z = 0 + 30 (75) = 2250
The maximum value of Z is 2400 and it occurs at R (30, 60)
Thus 30 units of A and 60 units of B must be manufactured to get maximum profit of ₹ 2400.
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