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WMA12/01 Jun 2025, Question 9: Stationary Points, Increasing and Optimising

Pearson Edexcel International AS Level Mathematics · Pure Mathematics 2 (WMA12) · Calculator · 11 marks

The question, in short

Open triangular-prism container with right-triangle ends, AE=3x, DE=2x, length L, capacity 12 m³. Find metal area formula S, minimise S, and justify the minimum.

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

Part (a) · 4 marks

Show the metal surface area S equals Px² + Q/x and find the positive integers P and Q.

  1. 1Each triangular end has area ½ × 3x × 2x = 3x²
  2. 2Volume = 3x² × L = 12, so L = 4/x²
  3. 3S = 2 × 3x² + 3x × L + 2x × L = 6x² + 5xL
  4. 4S = 6x² + 5x × 4/x² = 6x² + 20/x, so P = 6 and Q = 20

AnswerShown: S = 6x² + 20/x (P = 6, Q = 20)

Part (b) · 5 marks

Using calculus, find the value of x giving minimum S, and state minimum S to one decimal place.

  1. 1dS/dx = 12x − 20/x²
  2. 2Set dS/dx = 0: 12x = 20/x², so x³ = 5/3
  3. 3x = (5/3)^(1/3) = 1.18563…
  4. 4S = 6(1.18563)² + 20/1.18563 = 25.3

AnswerS = 25.3 m²

Part (c) · 2 marks

Justify that the S value found in part (b) is a minimum.

  1. 1d²S/dx² = 12 + 40/x³
  2. 2At x = 1.18563…, d²S/dx² = 12 + 40/1.6667 = 36 > 0
  3. 3So S = 25.3 is a minimum

AnswerShown: d²S/dx² = 12 + 40/x³ > 0 at x = 1.186, so S is a minimum

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Also searched as: WMA12 June 2025 · WMA12/01 · wma12 jun25 · IAS Maths Pure Mathematics 2 June 2025 · question 9.
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