2018 MI Promo Q11

The diagram shows an empty double-walled drinking glass. Both the inner and outer walls are made of ceramic glass in the shape of an open cylinder with a rim of \(0.2\)cm, which is also made of ceramic glass and there is a vacuum in between the two walls. The inner cylinder wall is of height \(h\) cm and radius \(r\) cm and the outer cylinder wall is of height \(0.2\)cm taller than that of the inner wall. Its total inner capacity is \(150\pi \) cm\(^{3}\).

It is also given that the ceramic glass is of negligible thickness.

1. Show that the surface area of the rim is equal to \(\frac{\pi }{25}\left( 10r+1 \right)\).[1]
2. Show that the total surface area of a drinking glass \(A\) is given by
   \(A=2\pi \left[ {{r}^{2}}+\frac{3}{5}r+\frac{2}{25}+\frac{300}{r}+\frac{30}{{{r}^{2}}} \right]\).[3]
3. Use differentiation to find the minimum value of \(A\).[4]

It is now given that \(h=2r\). One of such drinking glasses is filled to the brim with water and it is found to be leaking from the bottom of the glass. At time \(t\) seconds, the depth of water in the glass is \(x\)cm. It is assumed that the volume of the water in the glass is decreasing at a constant rate of \(0.5\)cm\(^{3}\) per second.

4. Calculate the rate at which the depth of water is decreasing.[4]