Maths · Calculus · chapter 6 of 13
Class 12 Maths Chapter 6: Application of Derivatives case study question with answers
One case-based question of the kind the board asks: a passage, then 3 parts worth 4 marks in all, with the answer and the marking scheme for each. Written from the NCERT chapter.
Case study · 4 marks
A small unit in Ludhiana stitches school bags. Its accountant finds that the total cost of making x bags a week is C(x) = 0.05x2 + 20x + 1500 rupees, and that to sell x bags a week the price per bag must be ₹ (80 − 0.05x), so the weekly revenue is R(x) = 80x − 0.05x2 rupees. The owner wants to know how the cost and the revenue change as output grows, and what weekly output earns the most.
Part 1 · 1 mark
The marginal cost when 100 bags are made is
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Answer: (a) ₹ 30
MC = C′(x) = 0.1x + 20 = 10 + 20 = ₹ 30. ₹ 25 halves the x term; ₹ 4000 is C(100).
Part 2 · 1 mark
The marginal revenue when 200 bags are sold is
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Answer: (b) ₹ 60
MR = R′(x) = 80 − 0.1x = 80 − 20 = ₹ 60. ₹ 70 halves the x term.
Part 3 · 2 marks
Find the weekly output that gives the maximum profit, and the maximum profit.
Answer and marking scheme
P(x) = R(x) − C(x) = 60x − 0.1x2 − 1500. P′(x) = 60 − 0.2x = 0 at x = 300; P″(x) = −0.2 < 0, a maximum. P(300) = 18000 − 9000 − 1500 = ₹ 7500.
- P(x) = 60x − 0.1x² − 1500 and P′(x) = 0 at x = 300 with P″ < 0 (1)
- maximum profit ₹ 7500 (1)
Or
Find the output for which marginal revenue equals marginal cost, and show that the profit increases for every output below it.
Answer and marking scheme
MR = MC: 80 − 0.1x = 0.1x + 20, so x = 300. P′(x) = MR − MC = 60 − 0.2x > 0 for 0 ≤ x < 300, so P is increasing there.
- 80 − 0.1x = 0.1x + 20 gives x = 300 (1)
- P′(x) = 60 − 0.2x > 0 for x < 300, so P is increasing (1)
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