2024 IGCSE Additional Mathematics February/March 0606/22 Q9

2024 IGCSE Additional Mathematics February/March 0606/22 Q9

9 marks

In this question all lengths are in centimetres and all angles are in radians.

The diagram shows a company logo. Each part of the logo is a sector of a circle with centre \(O\).

Sector \(AOB\) has radius \(x\).

Sector \(COD\) has radius \(x+2\).

Sector \(EOF\) has radius \(y\).

The shaded region has area \(A\,\mathrm{cm}^2\) and perimeter \(24\).

It is given that \(x\) and \(y\) can vary.

  1. Show that \(A=\dfrac{91}{8}x^2-68x+132\).[4]
  2. Use differentiation to find the minimum possible area of the logo.[5]

Solution:

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Answer:(a) \(A=\dfrac{91}{8}x^2-68x+132\). (b) \(\dfrac{2764}{91}\,\mathrm{cm}^2\approx30.4\,\mathrm{cm}^2\).

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