2026 CJC Promo Q10

2026 CJC Promo Q10

Junior College 1
12 marks
Promo

A metal can, in the shape of a closed cylinder of negligible thickness, has a fixed volume of \(K\text{ cm}^3\). The height and base radius of the can is \(h\text{ cm}\) and \(r\text{ cm}\) respectively, and neither exceeds \(8\text{ cm}\). The manufacturer of the metal cans wants to reduce the cost of production by minimizing the amount of metal used.

  1. Show that the total external surface area \(A\) of the can is given by \(A=2\left(\dfrac Kr+\pi r^2\right)\).[2]
  2. Use differentiation to find the value of \(r\) that minimizes \(A\). Give your answer in terms of \(K\).[4]
  3. Find the ratio \(h:r\) when \(A\) is minimum.[2]
  4. Deduce if the manufacturer is able to minimize \(A\) when the volume of the can is fixed at \(250\pi\text{ cm}^3\).[2]
  5. Sketch the graph showing the total external surface area of the can as the base radius varies, when the volume is fixed at \(100\text{ cm}^3\).[2]

Solution:

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Answer:(a) \(A=2\left(\dfrac Kr+\pi r^2\right)\). (b) Stationary minimum: \(r=\sqrt[3]{\dfrac K{2\pi}}\), if feasible. (c) \(2:1\). (d) The stationary minimum is infeasible; the constrained minimum occurs at \(h=8,r=\dfrac{5\sqrt5}{2}\). (e) Endpoints \((1.99,125),(8,427)\); minimum \((2.52,119)\).

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