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Date May 2021 Marks available 2 Reference code 21M.2.SL.TZ2.5
Level Standard Level Paper Paper 2 Time zone Time zone 2
Command term Find Question number 5 Adapted from N/A

Question

A hollow chocolate box is manufactured in the form of a right prism with a regular hexagonal base. The height of the prism is hcm, and the top and base of the prism have sides of length xcm.

Given that sin60°=32, show that the area of the base of the box is equal to 33x22.

[2]
a.

Given that the total external surface area of the box is 1200cm2, show that the volume of the box may be expressed as V=3003x-94x3.

[5]
b.

Sketch the graph of V=3003x-94x3, for 0x16.

[2]
c.

Find an expression for dVdx.

[2]
d.

Find the value of x which maximizes the volume of the box.

[2]
e.

Hence, or otherwise, find the maximum possible volume of the box.

[2]
f.

The box will contain spherical chocolates. The production manager assumes that they can calculate the exact number of chocolates in each box by dividing the volume of the box by the volume of a single chocolate and then rounding down to the nearest integer.

Explain why the production manager is incorrect.

[1]
g.

Markscheme

evidence of splitting diagram into equilateral triangles                M1

area =612x2sin60°               A1

=33x22               AG


Note: The AG line must be seen for the final A1 to be awarded.


[2 marks]

a.

total surface area of prism 1200=23x232+6xh               M1A1


Note: Award M1 for expressing total surface areas as a sum of areas of rectangles and hexagons, and A1 for a correctly substituted formula, equated to 1200.


h=400-3x22x               A1

volume of prism =332x2×h               (M1)

=332x2400-3x22x               A1

=3003x-94x3               AG


Note: The AG line must be seen for the final A1 to be awarded.


[5 marks]

b.

               A1A1

Note: Award A1 for correct shape, A1 for roots in correct place with some indication of scale (indicated by a labelled point).


[2 marks]

c.

dVdx=3003-274x2               A1A1


Note:
Award A1 for a correct term.


[2 marks]

d.

from the graph of V or dVdx  OR  solving dVdx=0               (M1)

x=8.88  8.877382               A1


[2 marks]

e.

from the graph of V  OR  substituting their value for x into V            (M1)

Vmax=3040cm3  3039.34               A1


[2 marks]

f.

EITHER
wasted space / spheres do not pack densely (tesselate)             A1

OR
the model uses exterior values / assumes infinite thinness of materials and hence the modelled volume is not the true volume             A1


[1 mark]

g.

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.
[N/A]
g.

Syllabus sections

Topic 5—Calculus » SL 5.3—Differentiating polynomials, n E Z
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Topic 5—Calculus

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