2025 DHS P1 Q11

[The surface area and volume of a sphere are given by \(4\pi r^2\) and \(\frac{4}{3}\pi r^3\) respectively.]

An aquarist who has a fixed budget of \$\(k\) wants to make a goldfish sanctuary. His design of the tank consists of a circular cover, a cylinder of length \(L\) cm and a hemispherical base of radius \(r\) cm neatly fitted together. The costs of the material per cm\(^2\) used for the cover, the cylindrical surface and hemispherical base are \$\(20\), \$\(10\) and \$\(30\) respectively. Assume that the tank is made of material with negligible thickness.

1. Show that the volume of the tank is \(V\) cm\(^3\), where \(V=\frac{k}{20}r-\frac{10}{3}\pi r^3\).

   [3]
2. As \(r\) varies, find the cost of the material used to make the hemispherical base in terms of \(k\) when \(V\) is at its maximum value. You need to show that \(V\) is a maximum.[3]

The aquarist has fixed the radius of the cylinder to be \(15\) cm. Initially, there is some water in the hemispherical part of the tank. However, due to a defect in the tank, water is leaking at a constant rate of \(20\) cm\(^3\) per minute.

The depth of the water at the hemispherical bottom is denoted by \(h\) cm, and the radius of the water surface is \(x\) cm. The volume of water in the hemispherical bottom is given by \(W\) cm\(^3\), where \(W=\frac{\pi h(3x^2+h^2)}{6}\)

3. By formulating a relationship between \(x\) and \(h\), find, when \(h=3\), the rate of change of
   1. the depth of water,[3]
   2. the radius of the water surface.[3]