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4PM1/02R Jan 2023, Question 11: Tangents, Normals and Stationary Points (Further)

Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 2R · Calculator · 10 marks

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The question, in short

Open cylinder radius r, height h, surface area 625π cm². Show h formula, then maximise volume V using calculus, justify maximum, and find h at that r.

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

Part (a) · 3 marks

Show that h equals (625 − r²) divided by 2r, using the open cylinder surface area 625π cm².

  1. 1Open cylinder: S = πr² + 2πrh
  2. 2πr² + 2πrh = 625π
  3. 32rh = 625 − r²
  4. 4h = (625 − r²)/(2r)

Answerh = (625 − r²)/(2r) shown

Part (b) · 6 marks

Use calculus to find r giving maximum V to 3 significant figures, and justify it is a maximum.

  1. 1V = πr² × (625 − r²)/(2r) = π(625r/2 − r³/2)
  2. 2dV/dr = 625π/2 − 3πr²/2
  3. 3dV/dr = 0 → 3r² = 625 → r = 14.4337… ≈ 14.4 (r > 0)
  4. 4d²V/dr² = −3πr < 0 for r > 0, so V is a maximum

Answerr = 14.4 cm; d²V/dr² = −3πr < 0 so maximum

Part (c) · 1 mark

Using the r from part (b), find the corresponding h to 3 significant figures.

  1. 1h = (625 − 14.4337…²)/(2 × 14.4337…)
  2. 2h = 14.4 cm

Answerh = 14.4 cm

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