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Sum
A firm manufactures 3 products A, B and C. The profits are Rs 3, Rs 2 and Rs 4 respectively. The firm has 2 machines and below is the required processing time in minutes for each machine on each product :
Machine | Products | ||
A | B | C | |
M1 M2 |
4 | 3 | 5 |
2 | 2 | 4 |
Machines M1 and M2 have 2000 and 2500 machine minutes respectively. The firm must manufacture 100 A's, 200 B's and 50 C's but not more than 150 A's. Set up a LPP to maximize the profit.
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Solution
Let the number of units of product A, B and C manufactured be x, y and z respectively.
Given, machine \[M_1\] takes 4 minutes to manufacture 1 unit of product A, 3 minutes to manufacture one unit of product B and 5 minute to manufacture one unit of product C.
Machine \[M_2\] takes 2 minutes to manufacture 1 unit of product A, 2 minutes to manufacture one unit of product B and 4 minute to manufacture one unit of product C.
The availability is 2000 minutes for
\[M_1\] and 2500 minutes for \[M_2\]
Thus,
\[4x + 3y + 5z \leq 2000\]
\[2x + 2y + 4z \leq 2500\]
\[2x + 2y + 4z \leq 2500\]
Number of units of products cannot be negative.
So,
So,
\[x, y, z \geq 0\]
Further, it is given that the firm should manufacture 100 A's, 200 B's and 50 C's but not more than 150 A's.
Then,
Then,
\[100 \leq x \leq 150\]
\[y \geq 200\]
\[z \geq 50\]
\[y \geq 200\]
\[z \geq 50\]
Let Z denotes the profit \[\therefore Z =\] 3x + 2y + 4z
Hence, the required LPP is as follows :
Maximize Z = 3x + 2y + 4z
Hence, the required LPP is as follows :
Maximize Z = 3x + 2y + 4z
subject to
\[4x + 3y + 5z \leq 2000\]
\[2x + 2y + 4z \leq 2500\]
\[2x + 2y + 4z \leq 2500\]
\[100 \leq x \leq 150\]
\[y \geq 200\]
\[z \geq 50\]
\[x, y, z \geq 0\]
\[y \geq 200\]
\[z \geq 50\]
\[x, y, z \geq 0\]
Concept: Mathematical Formulation of Linear Programming Problem
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