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Question
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 units 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 manufactured per month to maximize profit? How much is the maximum profit?
Solution
Let the firm manufactures x units of item A and y units of item B.
Firm can make profit of ₹ 20 per unit of A and ₹ 30 per unit of B.
Hence, the total profit is z = ₹ (20x + 30y)
This is the objective function which is to be maximized. The constraints are as per the following table:
| Item A (x) | Item A (y) | Total supply | |
| Motor | 3 | 2 | 210 |
| Transformer | 2 | 4 | 300 |
From the table, the constraints are
3x + 2y ≤ 210, 2x + 4y ≤ 300
Since, number of items cannot be negative, x ≥ 0, y ≥ 0.
Hence, the mathematical formulation of given LPP is :
Maximize z = 20x + 30y, subject to
3x + 2y ≤ 210, 2x + 4y ≤ 300, x ≥ 0, y ≥ 0
We draw the lines AB and CD whose equations are
3x + 2y = 210 and 2x + 4y = 300 respectively.
| Line | Equation | Points on the X-axis | Points on the Y-axis | Sign | Region |
| AB | 3x + 2y = 210 | A(70, 0) | B(0, 150) | ≤ | origin side of line AB |
| CD | 2x + 4y = 300 | C(150, 0) | D(0, 75) | ≤ | origin side of line CD |
The feasible region is OAPDO which is shaded in the graph.
The vertices of the feasible region are O (0, 0), A (70, 0), P and D (0, 75).
P is the point of intersection of the lines
2x + 4y = 300 ....(1)
and 3x + 2y = 210 ....(2)
Multiplying equation (2) by 2, we get
6x + 4y = 420
Subtracting equation (1) from this equation, we get
∴ 4x = 120
∴ x = 30
Substituting x = 30 in (1), we get
2(30) + 4y = 300
∴ 4y = 240
∴ y = 60
∴ P is (30, 60)
The values of the objective function z = 20x + 30y at these vertices are
z(O) = 20(0) + 30(0) = 0 + 0 = 0
z(A) = 20(70) + 30(0) = 1400 + 0 = 1400
z(P) = 20(30) + 30(60) = 600 + 1800 = 2400
z(D) = 20(0) + 30(75) = 0 + 2250 = 2250
∴ z has the maximum value 2400 when x = 30 and y = 60
Hence, the firm should manufactured 30 units of item A and 60 units of item B to get the maximum profit of ₹ 2400.
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