Now, let x, y and z units be the required production of products A,B and C.
Given:
Profit on one unit of product A is = Rs 3
So, profit on x unit of product A = 3x
Profit on one unit of product B is = Rs 20
So, profit on x unit of product B = 2y
Profit on one unit of product C is = Rs 4
So, profit on x unit of product C = 4z
Let total profit be ‘U’
So, U = 3x + 2y + 4z
First Constraint:
Given:
One unit of product A requires 4minutes on machine, M1
One unit of product B requires 3minutes on machine, M1
One unit of product C requires 5minutes on machine, M1
So,
x unit of product A requires 4x minutes on machine, M1
y unit of product B requires 3y minutes on machine, M1
z unit of product C requires 5z minutes on machine, M1
Total minutes on M1 = 2000 minutes
i.e., 4x + 3y + 5z ≤ 2000
Second constraint:
Given:
One unit of product A requires 2minutes on machine, M2
One unit of product B requires 2minutes on machine, M2
One unit of product C requires 4minutes on machine, M2
So,
x unit of product A requires 2x minutes on machine, M2
y unit of product B requires 2y minutes on machine, M2
z unit of product C requires 4z minutes on machine, M2
Total minutes on M2 = 2500 minutes
i.e., 2x + 2y + 4z ≤ 2500
Other constraints:
Given:
Firm must manufacture 100A’s, 200B’s and 50C’s but not more than 150A’s.
100 ≤ x ≤ 150
y ≥ 200
z ≥ 50
Hence, the required mathematical formulation of linear programming is:
Maximize U = 3x + 2y + 4z
Subject to
4x + 3y + 5z ≤ 2000
2x + 2y + 4z ≤ 2500
100 ≤ x ≤ 150
y ≥ 200
z ≥ 50
Where, x, y, z ≥ 0
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