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]