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WMA12/01 Jun 2023, Question 7: Stationary Points, Increasing and Optimising

Pearson Edexcel International AS Level Mathematics · Pure Mathematics 2 (WMA12) · Calculator · 8 marks

The question, in short

River height H = (√t / 20)(20 + 6t − t²) + 17 for 0 ≤ t ≤ 7 has a stationary value at t = α; use calculus to find α and classify it.

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

Part (a) · 5 marks

Differentiate H with respect to t, set dH/dt = 0, and show the stationary condition rearranges to 5α² − 18α − 20 = 0.

  1. 1H = t^(½) + (3/10)t^(3/2) − (1/20)t^(5/2) + 17
  2. 2dH/dt = ½t^(−½) + (9/20)t^(½) − (1/8)t^(3/2)
  3. 3Stationary at t = α: ½α^(−½) + (9/20)α^(½) − (1/8)α^(3/2) = 0
  4. 4Multiply by 40√α: 20 + 18α − 5α² = 0
  5. 55α² − 18α − 20 = 0

AnswerShown: dH/dt = ½t^(−½) + (9/20)t^(½) − (1/8)t^(3/2) = 0; multiplying by −40√t gives 5α² − 18α − 20 = 0

Part (b) · 1 mark

Solve 5α² − 18α − 20 = 0 for the positive root α, giving the answer to 3 decimal places.

  1. 1α = (18 ± √(18² + 4 × 5 × 20))/10 = (9 ± √181)/5
  2. 2Positive root in 0 ≤ t ≤ 7: α = 4.491 (3 d.p.)

Answerα = 4.491

Part (c) · 2 marks

Find the second derivative of H and evaluate it at α to prove H is a maximum there.

  1. 1d²H/dt² = −¼t^(−3/2) + (9/40)t^(−½) − (3/16)t^(½)
  2. 2At t = 4.491: d²H/dt² ≈ −0.0263 + 0.1062 − 0.3974 = −0.317
  3. 3d²H/dt² < 0, so H is a maximum at t = α

AnswerShown: d²H/dt² = −¼t^(−3/2) + (9/40)t^(−½) − (3/16)t^(½) ≈ −0.317 < 0 at t = 4.491, so H is a maximum

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Also searched as: WMA12 June 2023 · WMA12/01 · wma12 jun23 · IAS Maths Pure Mathematics 2 June 2023 · question 7.
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