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4PM1/01 Jan 2023, Question 8: Tangents, Normals and Stationary Points (Further)

Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 1 · Calculator · 13 marks

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The question, in short

Badge made from two x by y rectangles and a sector of radius x with angle 0.5 radians; area 50 cm². Show the perimeter formula, minimise P using calculus, and find the minimum P in the form k√2.

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

Part (a) · 5 marks

Show that the badge perimeter simplifies to P = 2x + 100/x using the given area of 50 cm².

  1. 1Area = 2xy + ½x²(0.5) = 2xy + x²/4 = 50
  2. 2So y = (50 − x²/4)/(2x)
  3. 3P = 2x + 4y + 0.5x (arc length CG = 0.5x)
  4. 44y = (100 − x²/2)/x = 100/x − x/2
  5. 5P = 2.5x + 100/x − x/2 = 2x + 100/x

AnswerP = 2x + 100/x shown

Part (b) · 6 marks

Use calculus to find the exact x giving minimum P, and justify that this x gives a minimum.

  1. 1dP/dx = 2 − 100/x²
  2. 22 − 100/x² = 0, x² = 50
  3. 3x = √50 = 5√2 (x > 0)
  4. 4d²P/dx² = 200/x³
  5. 5At x = 5√2, d²P/dx² > 0, so P is a minimum

Answerx = 5√2; d²P/dx² = 200/x³ > 0, so minimum

Part (c) · 2 marks

Find the minimum value of P, giving it in the form k√2 with integer k.

  1. 1P = 2(5√2) + 100/(5√2)
  2. 2= 10√2 + 10√2 = 20√2

Answer20√2

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