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WMA12/01 Jan 2023, Question 2: Stationary Points, Increasing and Optimising

Pearson Edexcel International AS Level Mathematics · Pure Mathematics 2 (WMA12) · Calculator · 9 marks

The question, in short

Cuboid brick has width x cm, length 3x cm, height h cm and volume 972 cm³; derive its surface area formula, then use calculus to locate and confirm the minimum surface area.

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

Part (a) · 3 marks

Use the volume 972 cm³ to express h in terms of x, then show the surface area simplifies to S = 6x² + 2592/x.

  1. 1S = 2(3x·x) + 2(3x·h) + 2(x·h) = 6x² + 8xh
  2. 2Volume: 3x²h = 972 ⇒ h = 324/x²
  3. 3S = 6x² + 8x × 324/x²
  4. 4S = 6x² + 2592/x

AnswerShown: h = 324/x², so S = 6x² + 8x(324/x²) = 6x² + 2592/x

Part (b) · 1 mark

Differentiate S = 6x² + 2592/x with respect to x.

  1. 1S = 6x² + 2592x⁻¹
  2. 2dS/dx = 12x − 2592x⁻² = 12x − 2592/x²

AnswerdS/dx = 12x − 2592/x²

Part (c) · 2 marks

Set dS/dx equal to zero and solve for the positive value of x where S is stationary.

  1. 112x − 2592/x² = 0
  2. 212x³ = 2592 ⇒ x³ = 216
  3. 3x = 6

Answerx = 6

Part (d) · 2 marks

Find d²S/dx², evaluate it at the x from part (c), and show this confirms a minimum.

  1. 1d²S/dx² = 12 + 5184/x³
  2. 2At x = 6: 12 + 5184/216 = 12 + 24 = 36
  3. 336 > 0, so x = 6 gives the minimum value of S

Answerd²S/dx² = 12 + 5184/x³; at x = 6 it equals 36 > 0, so S is a minimum

Part (e) · 1 mark

Substitute the x value into S to calculate the minimum surface area.

  1. 1S = 6(6)² + 2592/6
  2. 2S = 216 + 432 = 648 cm²

Answer648 cm²

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Also searched as: WMA12 January 2023 · WMA12/01 · wma12 jan23 · IAS Maths Pure Mathematics 2 January 2023 · question 2.
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