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Question
A company manufactures two types of ladies dresses C and D. The raw material and labour available per day is given in the table.
| Resources | Dress C(x) | Dress D(y) | Max. availability |
| Raw material | 5 | 4 | 60 |
| Labour | 5 | 3 | 50 |
P is the profit, if P = 50x + 100y, solve this LPP to find x and y to get the maximum profit
Solution
| Resources | Dress C(x) | Dress D(y) | Max. availability |
| Raw material | 5 | 4 | 60 |
| Labour | 5 | 3 | 50 |
From the table, the constraints are
5x + 4y ≤ 60
5x + 3y ≤ 50
∴ Given problem can be formulated as follows:
Maximize P = 50x + 100y
Subject to 5x + 4y ≤ 60
5x + 3y ≤ 50, x ≥ 0, y ≥ 0
To draw the feasible region, construct table as follows:
| Inequality | 5x + 4y ≤ 60 | 5x + 3y ≤ 50 |
| Corresponding equation (of line) | 5x + 4y = 60 | 5x + 3y = 50 |
| Intersection of line with X-axis | (12, 0) | (10, 0) |
| Intersection of line with Y-axis | (0, 15) | `(0, 50/3)` |
| Region | Origin side | Origin side |
Shaded portion OABC is the feasible region whose vertices are O(0, 0), A(10, 0), B and C(0, 15).
B is the point of intersection of 5x + 3y = 50 and 5x + 4y = 60.
Solving the above equations, we get
x = 4, y = 10
∴ B ≡ (4, 10)
Here, the objective function is
P = 50x + 100y
∴ P at O(0, 0) = 50(0) + 100(0) = 0
P at A(10, 0) = 50(10) + 100(0) = 500
P at B(4, 10) = 50(4) + 100(10)
= 200 + 1000
= 1200
P at C(0, 15) = 50(0) + 100(15) = 1500
∴ Maximum value of P is 1500 at x = 0 and y = 15.
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