WMA12/01 Jun 2021, Question 9: Stationary Points, Increasing and Optimising
Pearson Edexcel International AS Level Mathematics · Pure Mathematics 2 (WMA12) · Calculator · 10 marks
The question, in short
Open-top square-based box has volume 250000 cm³; use l and h to derive its external surface area formula, locate the stationary point, and confirm it is a minimum.
Summarised in our own words. Use the official paper for the exact wording and any diagrams.
Part (a) · 3 marks
Show that the external surface area S, in cm², equals 250000/h + 2000√h.
- 1hl² = 250 000, so l = 500/√h
- 2Open top: S = l² + 4hl
- 3l² = 250 000/h and 4hl = 4h × 500/√h = 2000√h
- 4S = 250 000/h + 2000√h
AnswerShown: l = 500/√h, so S = l² + 4hl = 250 000/h + 2000√h
Part (b) · 5 marks
Differentiate S algebraically and show a stationary point occurs at h = 250ᵏ, finding the rational constant k.
- 1dS/dh = −250 000h⁻² + 1000h^(−1/2)
- 2Set dS/dh = 0: 1000h^(−1/2) = 250 000/h²
- 3h²/√h = 250, so h^(3/2) = 250
- 4h = 250^(2/3), so k = 2/3
Answerk = 2/3 (h = 250^(2/3) ≈ 39.7)
Part (c) · 2 marks
Use a second derivative to justify that this h gives the minimum external surface area.
- 1d²S/dh² = 500 000h⁻³ − 500h^(−3/2)
- 2At h = 250^(2/3) ≈ 39.7: d²S/dh² = 8 − 2 = 6
- 36 > 0, so the stationary point is a minimum
Answerd²S/dh² = 500 000/h³ − 500h^(−3/2) = 6 > 0 at h = 250^(2/3), so S is a minimum.
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