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WMA12/01 Jan 2022, Question 2: Stationary Points, Increasing and Optimising

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

The question, in short

For the curve y = 27x^(1/2) − x^(3/2) − 20 with x > 0, differentiate, locate the stationary point, then use the second derivative to classify it, showing full working.

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

Part (a) · 2 marks

Differentiate y = 27x^(1/2) − x^(3/2) − 20 with respect to x, writing every term in its simplest form.

  1. 1Bring each power down and reduce it by 1
  2. 227x^(1/2) → 27 × ½ x^(−1/2) = (27/2)x^(−1/2)
  3. 3x^(3/2) → (3/2)x^(1/2), and the constant −20 → 0
  4. 4dy/dx = (27/2)x^(−1/2) − (3/2)x^(1/2)

Answerdy/dx = (27/2)x^(−1/2) − (3/2)x^(1/2)

Part (b) · 4 marks

Using your derivative, work out the coordinates of the point on C where the gradient is zero.

  1. 1Set (27/2)x^(−1/2) − (3/2)x^(1/2) = 0
  2. 2Multiply by 2x^(1/2): 27 − 3x = 0, so x = 9
  3. 3y = 27 × 3 − 27 − 20 = 34
  4. 4Stationary point is (9, 34)

Answer(9, 34)

Part (c) · 2 marks

Work out d²y/dx² and use its value to state whether the stationary point is a maximum or minimum.

  1. 1d²y/dx² = −(27/4)x^(−3/2) − (3/4)x^(−1/2)
  2. 2At x = 9: −(27/4)(1/27) − (3/4)(1/3) = −¼ − ¼ = −½
  3. 3d²y/dx² < 0, so (9, 34) is a maximum

Answerd²y/dx² = −(27/4)x^(−3/2) − (3/4)x^(−1/2); at x = 9 it is −½ < 0, so maximum

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Also searched as: WMA12 January 2022 · WMA12/01 · wma12 jan22 · IAS Maths Pure Mathematics 2 January 2022 · question 2.
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