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.

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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

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.