2024 CJC Promo Q10

A cone-shaped cup is made of paper of negligible thickness to hold \(20\pi \) cm\(^{3}\) of liquid. The open inverted cone has radius \(r\) cm and height \(h\) cm as shown in the diagram above. The external surface area of the cup is denoted by \(A\) cm\(^{2}\).

The manufacturer wants to reduce the cost of production by minimizing the value of \(A\).

1. Find \(h\) in terms of \(r\).[1]
2. By considering \({{A}^{2}}\) or otherwise, show that \(A\frac{\mathrm{d}A}{\mathrm{d}r}={{\pi }^{2}}\left( 2{{r}^{3}}-\frac{3600}{{{r}^{3}}} \right)\).[3]
3. Find the exact value of \(r\) that gives the minimum value of \(A\), proving that \(A\) is a minimum. Find also the ratio of the radius to the height, \(\frac{r}{h}\), giving your answer in terms of \(k\sqrt{2}\), where \(k\) is a constant to be determined.[5]

The manufacturer decides to make paper cups at minimum external surface area using the ratio \(\frac{r}{h}\) found in part (c).

4. The cup is being filled completely with water. However, there is a small hole at the bottom of the cup that causes water to leak out at the rate of \(3\) cm\(^{3}\) per second. Find the rate of decrease of the depth of the water at the instant when the depth is \(2\) cm.[3]

[It is given that the volume of a circular cone with base radius \(r\) and height \(h\) is \(\frac{1}{3}\pi {{r}^{2}}h\), and the curved surface area is \(\pi rl\), where \(l\) is the slant height of the cone.]