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Date	November 2016	Marks available	1	Reference code	16N.2.SL.TZ0.T_6
Level	Standard Level	Paper	Paper 2	Time zone	Time zone 0
Command term	Express	Question number	T_6	Adapted from	N/A

Question

A water container is made in the shape of a cylinder with internal height $h$ cm and internal base radius $r$ cm.

N16/5/MATSD/SP2/ENG/TZ0/06

The water container has no top. The inner surfaces of the container are to be coated with a water-resistant material.

The volume of the water container is $0.5{\text{ }}{{\text{m}}^3}$ .

The water container is designed so that the area to be coated is minimized.

One can of water-resistant material coats a surface area of $2000{\text{ c}}{{\text{m}}^2}$ .

Write down a formula for $A$ , the surface area to be coated.

[2]

a.

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

[1]

b.

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

[1]

c.

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

[2]

d.

Find $\frac{{{\text{d}}A}}{{{\text{d}}r}}$ .

[3]

e.

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

[3]

f.

Find the value of this minimum area.

[2]

g.

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

[3]

h.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

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

a.

$500\,000$ (A1)

Notes: Units not required.

[1 mark]

b.

$500\,000 = \pi {r^2}h$ (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]

c.

$A = \pi {r^2} + 2\pi r\left( {\frac{{500\,000}}{{\pi {r^2}}}} \right)$ (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]

d.

$2\pi r - \frac{{{\text{1}}\,{\text{000}}\,{\text{000}}}}{{{r^2}}}$ (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]

e.

$2\pi r - \frac{{1\,000\,000}}{{{r^2}}} = 0$ (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]

f.

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

g.

$\frac{{27\,679.0 \ldots }}{{2000}}$ (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]

h.

Examiners report

[N/A]

a.

[N/A]

b.

[N/A]

c.

[N/A]

d.

[N/A]

e.

[N/A]

f.

[N/A]

g.

[N/A]

h.

Syllabus sections

Topic 3—Geometry and trigonometry » SL 3.1—3d space, volume, angles, midpoints

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