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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 m3.

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 cm2.

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

[2]
a.

Express this volume in cm3.

[1]
b.

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

[1]
c.

Show that A=πr2+1000000r.

[2]
d.

Find dAdr.

[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=) πr2+2πrh    (A1)(A1)

 

Note:     Award (A1) for either πr2 OR 2πrh seen. Award (A1) for two correct terms added together.

 

[2 marks]

a.

500000    (A1)

 

Notes:     Units not required.

 

[1 mark]

b.

500000=πr2h    (A1)(ft)

 

Notes:     Award (A1)(ft) for πr2h equating to their part (b).

Do not accept unless V=πr2h is explicitly defined as their part (b).

 

[1 mark]

c.

A=πr2+2πr(500000πr2)    (A1)(ft)(M1)

 

Note:     Award (A1)(ft) for their 500000πr2 seen.

Award (M1) for correctly substituting only 500000πr2 into a correct part (a).

Award (A1)(ft)(M1) for rearranging part (c) to πrh=500000r and substituting for πrh in expression for A.

 

A=πr2+1000000r    (AG)

 

Notes:     The conclusion, A=πr2+1000000r, must be consistent with their working seen for the (A1) to be awarded.

Accept 106 as equivalent to 1000000.

 

[2 marks]

d.

2πr1000000r2    (A1)(A1)(A1)

 

Note:     Award (A1) for 2πr, (A1) for 1r2 or r2, (A1) for 1000000.

 

[3 marks]

e.

2πr1000000r2=0    (M1)

 

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

 

r3=10000002π OR r=310000002π     (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+1000000(54.1926)    (M1)

 

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

 

=27700 (cm2) (27679.0)    (A1)(ft)(G2)

[2 marks]

g.

27679.02000    (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.
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b.
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c.
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d.
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e.
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f.
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g.
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h.

Syllabus sections

Topic 3—Geometry and trigonometry » SL 3.1—3d space, volume, angles, midpoints
Show 90 related questions
Topic 5—Calculus » SL 5.1—Introduction of differential calculus
Topic 5—Calculus » SL 5.3—Differentiating polynomials, n E Z
Topic 2—Functions » SL 2.4—Key features of graphs, intersections using technology
Topic 4—Statistics and probability » SL 4.5—Probability concepts, expected numbers
Topic 2—Functions » SL 2.5—Modelling functions
Topic 4—Statistics and probability » SL 4.6—Combined, mutually exclusive, conditional, independence, prob diagrams
Topic 5—Calculus » SL 5.6—Stationary points, local max and min
Topic 5—Calculus » SL 5.7—Optimisation
Topic 1—Number and algebra » SL 1.8—Use of technology to solve systems of linear equations and polynomial equations
Topic 1—Number and algebra
Topic 2—Functions
Topic 3—Geometry and trigonometry
Topic 4—Statistics and probability
Topic 5—Calculus
Prior learning

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