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0606/23 May/Jun 2026, Question 9: Volume of a Prism

Cambridge IGCSE Additional Mathematics 0606 · Paper 23 (Calculator) · Calculator · 12 marks

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The question, in short

Prism with pentagon cross-section: square side x topped by equilateral triangle side x, length y. Volume given as 250(1 + √3/4). Show the surface area formula, then minimise A.

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

Part (a) · 5 marks

Show that the total surface area A equals (2 + √3/2)x² + 1250/x, using the given volume to eliminate y.

  1. 1Cross-section area = x² + (√3/4)x² = x²(1 + √3/4)
  2. 2Volume: x²(1 + √3/4)y = 250(1 + √3/4) ⇒ y = 250/x²
  3. 3Perimeter of cross-section = 5x, so the rectangular faces have area 5xy
  4. 4A = 2x²(1 + √3/4) + 5xy = (2 + √3/2)x² + 5x × 250/x²
  5. 5A = (2 + √3/2)x² + 1250/x

AnswerA = (2 + √3/2)x² + 1250/x shown

Part (bi) · 5 marks

Treating x as variable, find the minimum value of the surface area A.

  1. 1dA/dx = (4 + √3)x − 1250/x²
  2. 2dA/dx = 0 ⇒ x³ = 1250/(4 + √3) = 218.07
  3. 3x = 6.019
  4. 4A = (2 + √3/2)(6.019)² + 1250/6.019 = 311.5 ≈ 312

Answer312 cm² (311.5)

Part (bii) · 2 marks

Use the second derivative d²A/dx² to show that the value found is a minimum.

  1. 1d²A/dx² = (4 + √3) + 2500/x³
  2. 2At x = 6.019: d²A/dx² ≈ 5.73 + 11.46 = 17.2
  3. 3d²A/dx² > 0, so A is a minimum

Answerd²A/dx² = (4 + √3) + 2500/x³ ≈ 17.2 > 0 at x = 6.02, so minimum

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Also searched as: 0606/23/M/J/26 · 0606_s26_qp_23 · 0606 s26 23 · question 9.
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