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:

Asked by Sakshi | 1 year ago |  178

##### Solution :-

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

Answered by Aaryan | 1 year ago

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