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Question
A firm manufactures two products, each of which must be processed through two departments, 1 and 2. The hourly requirements per unit for each product in each department, the weekly capacities in each department, selling price per unit, labour cost per unit, and raw material cost per unit are summarized as follows:
| Product A | Product B | Weekly capacity | |
| Department 1 | 3 | 2 | 130 |
| Department 2 | 4 | 6 | 260 |
| Selling price per unit | Rs 25 | Rs 30 | |
| Labour cost per unit | Rs 16 | Rs 20 | |
| Raw material cost per unit | Rs 4 | Rs 4 |
The problem is to determine the number of units to produce each product so as to maximize total contribution to profit. Formulate this as a LPP.
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Solution
Let x and y units of product A and B were manufactured respectively.
Labour cost per unit to manufacture product A and product B is Rs 16 and Rs 20 respectively.Therefore, labour cost for x and y units of product A and product B is Rs 16x and Rs 20y respectively.
Total labour cost to manufacture product A and product B is Rs (16x+20y)
Raw material cost per unit to manufacture product A and product B is Rs 4 and Rs 4 respectively.Therefore,raw material cost for x and y units of product A and product B is Rs 4x and Rs 4y respectively.
Total raw material cost to manufacture product A and product B is Rs (4x + 4y)
Hence, total cost price to manufacture product A and product B = Total labour cost + Total raw material cost
= 16x + 4x + 20y + 4y
= 20x + 24y
Selling price per unit for product A and product B is Rs 25 and Rs 30 respectively. Therefore, total selling price for product A and product B is Rs 25x and Rs 30y respectively.
Total selling price = 25x + 30y
∴ Total profit = Total selling price − Total cost price = 25x + 30y
-(20 x + 24y)
=5x + 6y
Let Z denote the total profit
Then, Z = 5x + 5y
One unit of product A and product B requires 3 hours and 2 hours respectively at department 1.Therefore, x units and y units of product A and product B
require 3x hours and 2y hours respectively.
The weekly capacity of department 1 is 130.
One unit of product A and B requires 4 hours and 6 hours respectively at department 2.Therefore, x units and y units of product A and product B require 4x hours and 6y hours respectively.
The weekly capacity of department 2 is 260.
\[\therefore 4x + 6y \leq 260\]
Units of products cannot be negative.Therefore,
Maximize Z = 5x + 6y
subject to
\[3x + 2y \leq 130, \]
\[ 4x + 6y \leq 260, \]
\[x \geq 0, y \geq 0\]
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