Question 1 OR (a)
Two machines, two products: monthly output, unit cost and total cost by matrices
6 marks
- Time to write
- About 12 minutes
Question
Question 1 OR (a) · 6 marksA manufacturer makes two products A and B, each processed on machines I and II until completion. Machine I can process either 30 units of product A or 12 units of product B per hour. Machine II can process 50 of product B or 25 units of product A per hour.
Determine the following using matrices:
Monthly output of both products A and B if machines I and II are operated for 9 and 8 hours per day respectively in a 6-day working week of 4 weeks to a month.
Per unit cost of production if the operating cost per hour of the two machines is ₹24,000 and ₹36,000 respectively.
Total cost of production.
Reworded from the university paper: names and figures differ, the task is the same. The original paper belongs to the University of Delhi. See DU's question bank.
Before you write
Read off the question and its marks
Read from the question
- DetermineFormula first, then the working, then the answer with its unit (₹, %, units).
Write it in this order
- Read the data; note exactly what is asked 1 min
- 1 Hours available and the equations in matrix form 1 mark 2 min
- 2 Inverse and monthly output, with check 2 marks 3 min
- 3 Per unit cost by matrix product 2 marks 3 min
- 4 Total cost 1 mark 2 min
- Check: arithmetic, units (₹, %), final answer underlined 1 min
Try it first. About 12 minutes for 6 marks. Write it out, then open the solution.
Solution
Hover or tap a marked figure, like 72,000, to see how it is made and where else it appears.
Hours available in a month. A month has 6 × 4 = 24 working days, so machine I runs 9 × 24 = 216 hours and machine II runs 8 × 24 = 192 hours.
Time each unit takes. Every unit of A and of B passes through both machines. Machine I handles 30 units of A or 12 units of B in an hour, so a unit of A needs 1/30 hour and a unit of B 1/12 hour on it; on machine II a unit of A needs 1/25 hour and a unit of B 1/50 hour.
Let x and y be the monthly outputs of A and B. Each machine's hours are used in full:
x/30 + y/12 = 216 (machine I) x/25 + y/50 = 192 (machine II)
In matrix form TX = H, with T = [[1/30, 1/12], [1/25, 1/50]], X = [x, y]ᵀ and H = [216, 192]ᵀ.
|T| = (1/30)(1/50) − (1/12)(1/25) = 1/1500 − 1/300 = −4/1500 = −1/375, which is not zero, so T has an inverse and X = T⁻¹H:
adj T = [[1/50, −1/12], [−1/25, 1/30]], T⁻¹ = (adj T)/|T| = −375 × adj T = [[−7.5, 31.25], [15, −12.5]].
x = −7.5 × 216 + 31.25 × 192 = −1,620 + 6,000 = 4,380 y = 15 × 216 − 12.5 × 192 = 3,240 − 2,400 = 840
Monthly output: 4,380 units of A and 840 units of B.
Check: machine I, 4,380/30 + 840/12 = 146 + 70 = 216 hours; machine II, 4,380/25 + 840/50 = 175.2 + 16.8 = 192 hours.
(ii) Cost of production per unit
A unit's cost is the hours it spends on each machine times that machine's hourly operating cost, a row-by-column product:
[cost of A, cost of B]ᵀ = [[1/30, 1/25], [1/12, 1/50]] × [24,000, 36,000]ᵀ
| Product | Machine I (₹) | Machine II (₹) | Cost per unit (₹) |
|---|---|---|---|
| A | 24,000/30 = 800 | 36,000/25 = 1,440 | 2,240 |
| B | 24,000/12 = 2,000 | 36,000/50 = 720 | 2,720 |
Per unit cost: ₹2,240 for A and ₹2,720 for B.
(iii) Total cost of production
[4,380 840] × [2,240, 2,720]ᵀ:
| Product | Units | Cost per unit (₹) | Cost (₹) |
|---|---|---|---|
| A | 4,380 | 2,240 | 98,11,200 |
| B | 840 | 2,720 | 22,84,800 |
| Total | 1,20,96,000 |
Total cost of production: ₹1,20,96,000 a month.
Check: the machines' running cost for the month is 216 × 24,000 + 192 × 36,000 = 51,84,000 + 69,12,000 = ₹1,20,96,000, the same figure, since every machine hour goes into the output.
Where the 6 marks sit
| # | Component | Marks |
|---|---|---|
| 1 | Hours available and the equations in matrix form | 1 |
| 2 | Inverse and monthly output, with check | 2 |
| 3 | Per unit cost by matrix product | 2 |
| 4 | Total cost | 1 |
| Total | 6 |
Report a problemopens an email to admin@getkunji.com with the question ID1402-Q1aORalready filled in.