Solve Linear Programming Questions
A producer manufactures 3 models (I, II and III) of a particular product. He uses 2 raw materials A and B of which 4000 and 6000 units respectively are obtainable. The raw materials per unit of 3 models are listed below.
Raw materials

I

II

III

A

2

3

5

B

4

2

7

The labour time for each unit of model I is two times that of model II and thrice that of model III. The whole labour force of factory can manufacture an equivalent of 2500 units of model I. A model survey specifies that the minimum demand of 3 models is 500, 500 and 375 units correspondingly. However the ratio of number of units manufactured must be equal to 3:2:5. Suppose that gains per unit of model are 60, 40 and 100 correspondingly. Develop a LPP.
Answer
Assume
x_{1 } number of units of model I
x_{2 } number of units of model II
x_{3 } number of units of model III
Raw materials

I

II

III

Availability

A

2

3

5

4000

B

4

2

7

6000

Profit

60

40

100


x_{1} + 1/2x_{2} + 1/3x_{3} ≤ 2500 Labour time
x_{1} ≥ 500, x_{2} ≥ 500, x_{3} ≥ 375 Minimum demand
The given ratio is x_{1}: x_{2}: x_{3} = 3: 2: 5
x_{1 }/ 3 = x_{2 }/ 2 = x_{3 }/ 5 = k
x_{1 }= 3k; x_{2 }= 2k; x_{3 }= 5k
x_{2 }= 2k → k = x_{2 }/ 2
So x_{1 }= 3 x_{2 }/ 2 → 2x_{1} = 3x_{2}
Likewise 2x_{3 }= 5x_{2}
Maximize Z= 60x_{1} + 40x_{2} + 100x_{3}
Subject to 2x_{1} + 3x_{2} + 5x_{3} ≤ 4000
4x_{1} + 2x_{2} + 7x_{3} ≤ 6000
x_{1} + 1/2x_{2} + 1/3x_{3} ≤ 2500
2 x_{1} = 3x_{2}
2 x_{3 }= 5x_{2}
& x_{1} ≥ 500, x_{2} ≥ 500, x_{3} ≥ 375