A company produces two different products. One of them needs 1/4 of an hour of assembly work per unit, 1/8 of an hour in quality control work and Rs 1.2 in raw materials. The other product requires 1/3 of an hour of assembly work per unit, 1/3 of an hour in quality control work and Rs 0.9 in raw materials. Given the current availability of staff in the company, each day there is at most a total of 90 hours available for assembly and 80 hours for quality control. The first product described has a market value (sale price) of Rs 9 per unit and the second product described has a market value (sale price) of Rs 8 per unit. In addition, the maximum amount of daily sales for the first product is estimated to be 200 units, without there being a maximum limit of daily sales for the second product. Formulate and solve graphically the LPP and find the maximum profit.

This is a question of CBSE Sample Paper - Class 12 - 2017/18.

Question 29
A company produces two different products. One of them needs 1/4 of an hour of assembly work per unit, 1/8 of an hour in quality control work and Rs 1.2 in raw materials. The other product requires 1/3 of an hour of assembly work per unit, 1/3 of an hour in quality control work and Rs 0.9 in raw materials. Given the current availability of staff in the company, each day there is at most a total of 90 hours available for assembly and 80 hours for quality control. The first product described has a market value (sale price) of Rs 9 per unit and the second product described has a market value (sale price) of Rs 8 per unit. In addition, the maximum amount of daily sales for the first product is estimated to be 200 units, without there being a maximum limit of daily sales for the second product. Formulate and solve graphically the LPP and find the maximum profit.
Let Number of units of Product I produced be x,
Number of units of Product II produced be y
We need to maximize the Profit
Cost of Product I = Rs 1.2
Sale Price of Product I = Rs 9
Profit of Product I = Rs (9 – 1.2)
= Rs 7.8
Cost of Product II = Rs 0.9
Sale Price of Product II = Rs 8
Profit of Product II = Rs (8 – 0.9)
= Rs 7.1
∴ Z = 7.8x + 7.1y
Assembly Work
Time required on
Product I → 1/4 hour
Product II → 1/3 hour
Maximum Available Time = 90 hours
∴ 𝑥/4 + 𝑦/3≤ 90
(3𝑥 + 4𝑦)/12 ≤ 90
3x + 4y ≤ 1080
& x ≥ 0, y ≥ 0
Quality Control Work
Time required on
Product I → 1/8 hour
Product II → 1/3 hour
Maximum Available Time = 80 hours
∴ 𝑥/8 + 𝑦/3≤ 80
(3𝑥 + 8𝑦)/24 ≤ 80
3x + 8y ≤ 1920
& x ≥ 0, y ≥ 0
Also,
Maximum sales of first product is 200 units
∴ x ≤ 200
Combining all constraints :
Max Z = 7.8x + 7.1y
Subject to constraints,
3x + 4y ≤ 1080,
3x + 8y ≤ 1920
x ≤ 200,
& x ≥ 0 , y ≥ 0
Hence, profit will be maximum if
Number of Product I = 200
Number of Product II = 120
Maximum Profit = Rs. 2412

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.