Write down a formula for \(A\), the surface area to be coated.

Express this volume in \({\text{c}}{{\text{m}}^3}\).

Write down, in terms of \(r\) and \(h\), an equation for the volume of this water container.

Show that \(A = \pi {r^2}\frac{{1\,000\,000}}{r}\).

Show that \(A = \pi {r^2} + \frac{{1\,000\,000}}{r}\).

Find \(\frac{{{\text{d}}A}}{{{\text{d}}r}}\).

Using your answer to part (e), find the value of \(r\) which minimizes \(A\).

Find the value of this minimum area.

Find the least number of cans of water-resistant material that will coat the area in part (g).

\((A = ){\text{ }}\pi {r^2} + 2\pi rh\)    (A1)(A1)

Note:     Award (A1) for either \(\pi {r^2}\) OR \(2\pi rh\) seen. Award (A1) for two correct terms added together.

[2 marks]

\(500\,000\)    (A1)

Notes:     Units not required.

[1 mark]

\(500\,000 = \pi {r^2}h\)    (A1)(ft)

Notes:     Award (A1)(ft) for \(\pi {r^2}h\) equating to their part (b).

Do not accept unless \(V = \pi {r^2}h\) is explicitly defined as their part (b).

[1 mark]

\(A = \pi {r^2} + 2\pi r\left( {\frac{{500\,000}}{{\pi {r^2}}}} \right)\)    (A1)(ft)(M1)

Note:     Award (A1)(ft) for their \({\frac{{500\,000}}{{\pi {r^2}}}}\) seen.

Award (M1) for correctly substituting only \({\frac{{500\,000}}{{\pi {r^2}}}}\) into a correct part (a).

Award (A1)(ft)(M1) for rearranging part (c) to \(\pi rh = \frac{{500\,000}}{r}\) and substituting for \(\pi rh\) in expression for \(A\).

\(A = \pi {r^2} + \frac{{1\,000\,000}}{r}\)    (AG)

Notes:     The conclusion, \(A = \pi {r^2} + \frac{{1\,000\,000}}{r}\), must be consistent with their working seen for the (A1) to be awarded.

Accept \({10^6}\) as equivalent to \({1\,000\,000}\).

[2 marks]

\(A = \pi {r^2} + 2\pi r\left( {\frac{{500\,000}}{{\pi {r^2}}}} \right)\)    (A1)(ft)(M1)

Note:     Award (A1)(ft) for their \({\frac{{500\,000}}{{\pi {r^2}}}}\) seen.

Award (M1) for correctly substituting only \({\frac{{500\,000}}{{\pi {r^2}}}}\) into a correct part (a).

Award (A1)(ft)(M1) for rearranging part (c) to \(\pi rh = \frac{{500\,000}}{r}\) and substituting for \(\pi rh\) in expression for \(A\).

\(A = \pi {r^2} + \frac{{1\,000\,000}}{r}\)    (AG)

Notes:     The conclusion, \(A = \pi {r^2} + \frac{{1\,000\,000}}{r}\), must be consistent with their working seen for the (A1) to be awarded.

Accept \({10^6}\) as equivalent to \({1\,000\,000}\).

[2 marks]

\(2\pi r - \frac{{{\text{1}}\,{\text{000}}\,{\text{000}}}}{{{r^2}}}\)    (A1)(A1)(A1)

Note:     Award (A1) for \(2\pi r\), (A1) for \(\frac{1}{{{r^2}}}\) or \({r^{ - 2}}\), (A1) for \( - {\text{1}}\,{\text{000}}\,{\text{000}}\).

[3 marks]

\(2\pi r - \frac{{1\,000\,000}}{{{r^2}}} = 0\)    (M1)

Note:     Award (M1) for equating their part (e) to zero.

\({r^3} = \frac{{1\,000\,000}}{{2\pi }}\) OR \(r = \sqrt[3]{{\frac{{1\,000\,000}}{{2\pi }}}}\)     (M1)

Note:     Award (M1) for isolating \(r\).

OR

sketch of derivative function     (M1)

with its zero indicated     (M1)

\((r = ){\text{ }}54.2{\text{ }}({\text{cm}}){\text{ }}(54.1926 \ldots )\)    (A1)(ft)(G2)

[3 marks]

\(\pi {(54.1926 \ldots )^2} + \frac{{1\,000\,000}}{{(54.1926 \ldots )}}\)    (M1)

Note:     Award (M1) for correct substitution of their part (f) into the given equation.

\( = 27\,700{\text{ }}({\text{c}}{{\text{m}}^2}){\text{ }}(27\,679.0 \ldots )\)    (A1)(ft)(G2)

[2 marks]

\(\frac{{27\,679.0 \ldots }}{{2000}}\)    (M1)

Note:     Award (M1) for dividing their part (g) by 2000.

\( = 13.8395 \ldots \)    (A1)(ft)

Notes:     Follow through from part (g).

14 (cans)     (A1)(ft)(G3)

Notes:     Final (A1) awarded for rounding up their \(13.8395 \ldots \) to the next integer.

[3 marks]