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4PM1/02R Jun 2022, Question 11: Tangents, Normals and Stationary Points (Further)

Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 2R · Calculator · 12 marks

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The question, in short

Cylinder radius r, height h, total surface area 600 cm², volume V. Show the given volume formula, find r maximising V using calculus, justify the maximum, then find the largest sphere radius p after melting.

Summarised in our own words. Use the official paper for the exact wording and the diagram.

Part (a) · 4 marks

Using total surface area 600 cm², eliminate h and show that the volume simplifies to V = 300r − πr³.

  1. 12πr² + 2πrh = 600, so πr² + πrh = 300
  2. 2h = (300 − πr²)/(πr)
  3. 3V = πr²h = πr² × (300 − πr²)/(πr)
  4. 4V = r(300 − πr²) = 300r − πr³

AnswerV = 300r − πr³ shown

Part (b) · 5 marks

Differentiate V with respect to r, solve for the stationary r = √(100/π), and justify it gives a maximum.

  1. 1dV/dr = 300 − 3πr²
  2. 2300 − 3πr² = 0, r² = 100/π
  3. 3r = √(100/π) (r > 0)
  4. 4d²V/dr² = −6πr
  5. 5At r = √(100/π), d²V/dr² = −6π√(100/π) ≈ −106.3 < 0, so V is a maximum

Answer(i) dV/dr = 300 − 3πr² = 0 gives r = √(100/π) — shown (ii) d²V/dr² = −6πr < 0 for r > 0, so maximum

Part (c) · 3 marks

Equate cylinder volume at maximum to sphere volume, then find p to one decimal place.

  1. 1Maximum V = 300√(100/π) − π(√(100/π))³ = 1128.38…
  2. 2(4/3)πp³ = 1128.38…
  3. 3p³ = 269.38…
  4. 4p = 6.5 cm

Answerp = 6.5

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