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Date November 2016 Marks available 2 Reference code 16N.2.SL.TZ0.T_6
Level Standard Level Paper Paper 2 Time zone Time zone 0
Command term Show that 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   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  c m 2 .

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

[2]
a.

Express this volume in  c 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 = π r 2 + 1 000 000 r .

[2]
d.

Find d A 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 = )   π r 2 + 2 π r h    (A1)(A1)

 

Note:     Award (A1) for either π r 2  OR 2 π r h seen. Award (A1) for two correct terms added together.

 

[2 marks]

a.

500 000    (A1)

 

Notes:     Units not required.

 

[1 mark]

b.

500 000 = π r 2 h    (A1)(ft)

 

Notes:     Award (A1)(ft) for π r 2 h equating to their part (b).

Do not accept unless V = π r 2 h is explicitly defined as their part (b).

 

[1 mark]

c.

A = π r 2 + 2 π r ( 500 000 π r 2 )    (A1)(ft)(M1)

 

Note:     Award (A1)(ft) for their 500 000 π r 2 seen.

Award (M1) for correctly substituting only 500 000 π r 2 into a correct part (a).

Award (A1)(ft)(M1) for rearranging part (c) to π r h = 500 000 r and substituting for π r h  in expression for A .

 

A = π r 2 + 1 000 000 r    (AG)

 

Notes:     The conclusion, A = π r 2 + 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 π r 1 000 000 r 2    (A1)(A1)(A1)

 

Note:     Award (A1) for 2 π r , (A1) for 1 r 2 or r 2 , (A1) for 1 000 000 .

 

[3 marks]

e.

2 π r 1 000 000 r 2 = 0    (M1)

 

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

 

r 3 = 1 000 000 2 π OR  r = 1 000 000 2 π 3     (M1)

 

Note:     Award (M1) for isolating r .

 

OR

sketch of derivative function     (M1)

with its zero indicated     (M1)

( r = )   54.2   ( cm )   ( 54.1926 )    (A1)(ft)(G2)

[3 marks]

f.

π ( 54.1926 ) 2 + 1 000 000 ( 54.1926 )    (M1)

 

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

 

= 27 700   ( c m 2 )   ( 27 679.0 )    (A1)(ft)(G2)

[2 marks]

g.

27 679.0 2000    (M1)

 

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

 

= 13.8395    (A1)(ft)

 

Notes:     Follow through from part (g).

 

14 (cans)     (A1)(ft)(G3)

 

Notes:     Final (A1) awarded for rounding up their 13.8395 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, distance, midpoints
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Topic 5 —Calculus » SL 5.1—Introduction of differential calculus
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Topic 4—Statistics and probability » SL 4.5—Probability concepts, expected numbers
Topic 5 —Calculus » SL 5.5—Integration introduction, areas between curve and x axis
Topic 4—Statistics and probability » SL 4.6—Combined, mutually exclusive, conditional, independence, prob diagrams
Topic 5 —Calculus » SL 5.8—Testing for max and min, optimisation. Points of inflexion
Topic 3— Geometry and trigonometry
Topic 4—Statistics and probability
Topic 5 —Calculus
Prior learning

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