- \hspace{0pt} [In this question you may assume the following formulae for the volume and curved] surface area of a cone of base radius \(r\) and height \(h\) and of a sphere of radius \(r\).
Cone: volume \(V = \frac { 1 } { 3 } \pi r ^ { 2 } h\) and curved surface area \(S = \pi r \sqrt { h ^ { 2 } + r ^ { 2 } }\)
Sphere: volume \(V = \frac { 4 \pi } { 3 } r ^ { 3 }\) and curved surface area \(S = 4 \pi r ^ { 2 }\)
\includegraphics[max width=\textwidth, alt={}, center]{78ba3acc-4cca-4d15-8362-a27e425c5859-22_782_755_637_657}
Figure 3
Figure 3 shows the design for a garden ornament.
The ornament is made of a hemisphere on top of a truncated cone.
The truncated cone has base radius \(2 r \mathrm {~cm}\), top radius \(r \mathrm {~cm}\) and height \(4 r \mathrm {~cm}\).
The hemisphere has radius \(R \mathrm {~cm}\).
Given that the volume of the ornament is \(2100 \pi \mathrm {~cm} ^ { 3 }\)
- show that
$$R ^ { 3 } = 3150 - 14 r ^ { 3 }$$
- Find an expression involving \(\frac { \mathrm { d } R } { \mathrm {~d} r }\) in terms of \(r\) and/or \(R\).
The base of the truncated cone of the ornament is fixed to the ground.
- Show that the visible surface area of the ornament, \(A \mathrm {~cm} ^ { 2 }\), is given by
$$A = ( 3 \sqrt { 17 } - 1 ) \pi r ^ { 2 } + 3 \pi R ^ { 2 }$$
- Hence show that
$$\frac { \mathrm { d } A } { \mathrm {~d} r } = \gamma \pi r - \frac { \delta \pi r ^ { 2 } } { R }$$
where \(\gamma\) and \(\delta\) are real numbers to be determined.
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{78ba3acc-4cca-4d15-8362-a27e425c5859-23_705_803_625_630}
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\caption{Figure 4}
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Figure 4 shows a sketch of \(A\) against \(r\), for \(r \geqslant 0\)
There is a local minimum at \(r = 0\) and a local maximum at the point \(M\). The overall minimum point is at the point \(N\), where the gradient of the curve is undefined. - Determine the \(r\) coordinate of the point \(N\).
- Explain why, for the ornament, \(r\) must be less than this value.
- Show that the \(r\) coordinate of the point \(M\) is
$$\sqrt [ 3 ] { \frac { p ( 3 \sqrt { 17 } - 1 ) ^ { 3 } } { 3 q ^ { 2 } + ( 3 \sqrt { 17 } - 1 ) ^ { 3 } } }$$
where \(p\) and \(q\) are integers to be determined.