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0606/22 Feb/Mar 2026, Question 7: Tangents, Normals and Stationary Points (Further)

Cambridge IGCSE Additional Mathematics 0606 · Paper 22 (Calculator) · Calculator · 8 marks

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The question, in short

A ship's hourly sailing cost is C = v² + 3000/v + 100 for speed v km h⁻¹. Determine the speed giving least cost, confirm it is a minimum, then find the least cost for a 150 km trip.

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Part (a) · 5 marks

Differentiate C with respect to v, solve for the speed where the gradient is zero, and use a suitable test to confirm the cost is minimised there.

  1. 1dC/dv = 2v − 3000/v²
  2. 22v − 3000/v² = 0 gives v³ = 1500
  3. 3v = ∛1500 = 11.4 km h⁻¹
  4. 4d²C/dv² = 2 + 6000/v³ > 0 when v > 0, so this is a minimum

Answerv = ∛1500 or 11.4 km h⁻¹; d²C/dv² = 2 + 6000/v³ > 0, so C is a minimum

Part (b) · 3 marks

Use your speed from part (a) to work out the total cost of covering 150 km.

  1. 1Time = 150/∛1500 hours = 13.10… h
  2. 2Cost per hour = (∛1500)² + 3000/∛1500 + 100 = 493.1…
  3. 3Total cost = 493.1… × 13.10… ≈ $6460

Answer$6460 to $6510

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