2025 YIJC Promo Q7

2025 YIJC Promo Q7

Junior College 1
6 marks

A piece of wire of fixed length \(k\) m is cut into two parts. One part is bent into the shape of a rectangle, with sides of length \(x\) m and \(y\) m. The other part is bent into the shape of an equilateral triangle, with sides of length \(x\) m.

  1. Show that the total area of the two shapes, \(A\) m\(^2\), can be expressed as \(A=\frac{k}{2}x+\frac{\sqrt{3}-10}{4}x^2\).[3]
  2. Use differentiation to find \(x\) in terms of \(k\) that maximises \(A\), showing that your answer gives a maximum \(A\).[3]

Solution:

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Answer:\(A=\dfrac{k}{2}x+\dfrac{\sqrt3-10}{4}x^2\); maximum at \(x=\dfrac{k}{10-\sqrt3}\)

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