Question 2 (b)
Replacement time of a pickup truck by calculus
9 marks
- Time to write
- About 18 minutes
Question
Question 2 (b) · 9 marksA transport company buys a pickup truck at a cost of ₹84,00,000. The company's estimate is that the average cost of capital and the average operating cost are a function of x, i.e., the number of days for which the truck is used. The truck's salvage value (in rupees) is given by the function (x) = 7536000 - 6000 x. The operating cost per day, OC, is given by the function OC(x) = 8000 + 0.3 x.
Determine the following using calculus:
The number of days for which the truck should be used before its replacement, if the objective is to minimize the sum of the average capital.
The average depreciation of the truck per day.
The average operating cost of the truck per day.
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
- estimateFormula first, then the working, then the answer with its unit (₹, %, units).
What the paper leaves open
As printed, the salvage value function is written without its symbol (the line reads "(x) = 7536000 - 6000 x"); it is kept as printed and read as the salvage value function. Part (i) is kept as printed ("the sum of average capital").
Write it in this order
- Read the data; note exactly what is asked 1 min
- 1 Average depreciation and operating cost functions 3 marks 5 min
- 2 Optimal days with second-order condition 3 marks 5 min
- 3 Average depreciation per day 1½ marks 3 min
- 4 Average operating cost per day 1½ marks 3 min
- Check: arithmetic, units (₹, %), final answer underlined 1 min
Try it first. About 18 minutes for 9 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.
Reading of the data. The salvage value after x days is S(x) = 75,36,000 − 6,000x (the function printed without its symbol, as the paper note says). OC(x) = 8,000 + 0.3x is read as the operating cost on the x-th day (it rises as the truck ages), so the total operating cost over x days is the integral of OC and the average is that total divided by x. Part (i) is read as minimising the sum of the average capital cost (depreciation) per day and the average operating cost per day.
Average capital cost (depreciation) per day
Capital consumed in x days = cost − salvage value = 84,00,000 − (75,36,000 − 6,000x) = 8,64,000 + 6,000x.
Average depreciation per day: AD(x) = (8,64,000 + 6,000x)/x = 8,64,000/x + 6,000.
Average operating cost per day
Total operating cost over x days = ∫₀ˣ (8,000 + 0.3t) dt = 8,000x + 0.15x².
Average operating cost per day: AO(x) = 8,000 + 0.15x.
Sum of averages: A(x) = 8,64,000/x + 6,000 + 8,000 + 0.15x = 8,64,000/x + 14,000 + 0.15x.
dA/dx = −8,64,000/x² + 0.15 = 0 ⟹ x² = 57,60,000 ⟹ x = 2,400 days.
d²A/dx² = 17,28,000/x³ > 0, so A is a minimum at x = 2,400 (second derivative test).
The truck should be used for 2,400 days before it is replaced.
(ii) Average depreciation per day at x = 2,400: 8,64,000/2,400 + 6,000 = 360 + 6,000 = ₹6,360.
(iii) Average operating cost per day at x = 2,400: 8,000 + 0.15 × 2,400 = ₹8,360.
| Average cost per day at x = 2,400 | ₹ |
|---|---|
| Depreciation | 6,360 |
| Operating cost | 8,360 |
| Total | 14,720 |
At the optimum the average depreciation falls as fast as the average operating cost rises (8,64,000/2,400² = 0.15 = the slope of AO).
Note on the salvage value. S(2,400) = 75,36,000 − 6,000 × 2,400 = −68,64,000. The linear salvage function given turns negative after 1,256 days, so it cannot be a market value over the whole period; it is used, as the question directs, only as the function that sets capital consumption. The optimum depends only on its slope (₹6,000 a day) and on the initial loss (cost − 75,36,000 = ₹8,64,000).
The other reading. If OC(x) = 8,000 + 0.3x were itself the average operating cost per day over x days, then A(x) = 8,64,000/x + 14,000 + 0.3x, x = √28,80,000 ≈ 1,697.06 days, average depreciation ≈ ₹6,509.12 and average operating cost ≈ ₹8,509.12 a day.
Where the 9 marks sit
| # | Component | Marks |
|---|---|---|
| 1 | Average depreciation and operating cost functions | 3 |
| 2 | Optimal days with second-order condition | 3 |
| 3 | Average depreciation per day | 1.5 |
| 4 | Average operating cost per day | 1.5 |
| Total | 9 |
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