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1 Pure Mathematics

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Question 387

A landscape garden feature is designed using two identical sectors of a circle, ORP ORP\,ORP and OQSOQSOQS, and a central rhombus-shaped flower bed OSTROSTROSTR, as shown in the layout. The points PPP, OOO, and Q Q\,Q lie on a straight stone boundary of length 12 metres, such that O O\,O is the midpoint of PQPQPQ. The radii OR OR\,OR and OS OS\,OS form two sides of the rhombus, and the angle ∠ROS \angle ROS\,∠ROS is denoted by θ \theta\,θ radians.

Layout of two identical sectors ORP and OQS beside rhombus OSTR, with P, O and Q collinear, PQ equal to 12 metres, and angle ROS labelled theta.

a.

Show that the total area of the garden feature, A A\,A square metres, is given by

A=18(π−θ+2sin⁡θ) A = 18(\pi - \theta + 2\sin\theta) A=18(π−θ+2sinθ)
[4]
bi.

Use calculus to show that the maximum value of A A\,A occurs when θ=π3\displaystyle \theta = \frac{\pi}{3}θ=3π​. Fully justify that this value of θ \theta\,θ gives a maximum.

[5]
bii.

Determine the exact maximum value of AAA.

[2]
c.

Without further calculation, state how your answers to parts (b)(i) and (b)(ii) would change if the total boundary length PQ PQ\,PQ were increased to 24 metres.

[2]
Markscheme

1 Pure Mathematics Questions

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