2020 TMJC Promo Q10

A manufacturer designs a cylindrical water dispenser with a fixed volume, \(V\)cm\(^{3}\) where \(V>1\). The manufacturer decides to use stainless steel, with negligible thickness, for the cylindrical body of the water dispenser of height \(y\)cm and the circular base of radius \(r\) cm. Plastic, with negligible thickness, will be used instead for the circular lid of radius \(r\) cm, as shown in Diagram 1. In order to insert a tap, an area of \(1\)cm\(^{2}\) of stainless steel will be removed from the cylindrical body of the water dispenser. The total area of stainless steel to be used for the cylindrical body and circular base of the water dispenser, as shown in Diagram 2, is denoted by \(A\)cm\(^{2}\). The manufacturer wants the value of \(A\) to be as small as possible.

1. Show that \(A=\pi {{r}^{2}}+\frac{2V}{r}-1\).[2]
2. Using differentiation, find in terms of \(V\) and \(\pi \), the minimum value of \(A\), proving that  it is a minimum.
   Simplify your answer.[4]
3. Sketch the graph of \(A\) as \(r\) varies, stating the coordinates of any turning point(s) in terms of \(V\)and \(\pi \)and the equations of any asymptotes.[3]

It is given that the water dispenser, with the minimum value of \(A\) found in part (ii), is completely filled with water. When the tap is turned on, the rate of decrease of the volume of water, \(W\)cm\(^{3}\), in the water dispenser with respect to time \(t\) seconds is \(y{{\mathrm{e}}^{-\frac{t}{\pi {{r}^{3}}}}}\)cm\(^{3}\)/s.

4. Find the rate of change of the height of water, \(h\) cm, in the water dispenser \(1\) minute after the tap is turned on, as shown in Diagram 3. Simplify your answer in terms of \(V\) and \(\pi \).[3]