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4PM1/02R Jun 2023, Question 3: 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

Sector OAB has radius r cm, angle θ radians and area 675 cm². Show the perimeter formula, then use calculus to find and justify the minimum perimeter.

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

Part (a) · 3 marks

Using the sector area of 675 cm², show that the perimeter P equals 2r + 1350/r.

  1. 1Area: (1/2)r²θ = 675, so θ = 1350/r²
  2. 2P = 2r + rθ
  3. 3P = 2r + r × 1350/r² = 2r + 1350/r

AnswerP = 2r + 1350/r shown

Part (b) · 5 marks

Differentiate P with respect to r to find the minimum value of P, giving it as a√b with a integer and b prime.

  1. 1dP/dr = 2 − 1350/r²
  2. 22 − 1350/r² = 0, so r² = 675, r = 15√3
  3. 3P = 2(15√3) + 1350/(15√3)
  4. 4= 30√3 + 30√3 = 60√3

Answer60√3

Part (c) · 2 marks

Give a reason, using calculus, that the value of P found in part b is a minimum.

  1. 1d²P/dr² = 2700/r³
  2. 2r > 0, so d²P/dr² > 0 (≈ 0.15 at r = 15√3)
  3. 3So the value is a minimum

Answerd²P/dr² = 2700/r³ > 0 for r > 0, so P is a minimum

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