O-LevelAdditional MathematicsDifferentiation and integration4037/21 · May/June 2024 · Q117 Marks

In this question all lengths are in centimetres. The diagram shows a rectangle ABCD with BC = x. The area of the rectangle is 400 cm². Two identical quarter-circles of radius x/2, with centres A and C, are removed from the rectangle to make the shaded shape. Given that x can vary, find the value of x that gives the minimum value of the perimeter of the shaded shape and hence find this minimum value.

In this question all lengths are in centimetres. The diagram shows a rectangle ABCD with BC = x. The area of the rectangle is 400 cm². Two identical quarter-circles of radius x/2, with centres A and C, are removed from the rectangle to make the shaded shape. Given that x can vary, find the value of x that gives the minimum value of the perimeter of the shaded shape and hence find this minimum value.[7]

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Mark scheme

Marking points from the official mark scheme (May/June 2024). The mark or mark type is shown in brackets.

[B1] rectangle length = 400/x soi
[B2] P = 2(π(x/4)) + x + 2(400/x - x/2) oe
[B1] dP/dx = π/2 - 800/x² oe (Correct first derivative)
[M1] Equates their dP/dx to 0 and solves for x
[A1] x = 40/√π or 22.6 or 22.56[75...]
[M1] Substitutes their x into their P (FT their value of x provided it is greater than 0)
[A1] P = 70.9 or 70.89[81...]

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