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Date November 2020 Marks available 3 Reference code 20N.2.SL.TZ0.T_4
Level Standard Level Paper Paper 2 Time zone Time zone 0
Command term Find Question number T_4 Adapted from N/A

Question

Hyungmin designs a concrete bird bath. The bird bath is supported by a pedestal. This is shown in the diagram.

The interior of the bird bath is in the shape of a cone with radius r, height h and a constant slant height of 50cm.

Let V be the volume of the bird bath.

Hyungmin wants the bird bath to have maximum volume.

Write down an equation in r and h that shows this information.

[1]
a.

Show that V=2500πh3-πh33.

[1]
b.

Find dVdh.

[2]
c.

Using your answer to part (c), find the value of h for which V is a maximum.

[2]
d.

Find the maximum volume of the bird bath.

[2]
e.

To prevent leaks, a sealant is applied to the interior surface of the bird bath.

Find the surface area to be covered by the sealant, given that the bird bath has maximum volume.

[3]
f.

Markscheme

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

h2+r2=502  (or equivalent)        (A1)


Note: Accept equivalent expressions such as r=2500-h2 or h=2500-r2. Award (A0) for a final answer of ±2500-h2 or ±2500-r2, or any further incorrect working.


[1 mark]

a.

13×π×2500-h2×h  OR  13×π×2500-h22×h        (M1)


Note: Award (M1) for correct substitution in the volume of cone formula.


V=2500πh3-πh33        (AG)


Note: The final line must be seen, with no incorrect working, for the (M1) to be awarded.


[1 mark]

b.

dVdh= 2500π3-πh2        (A1)(A1)


Note: Award (A1) for 2500π3, (A1) for -πh2. Award at most (A1)(A0) if extra terms are seen. Award (A0) for the term -3πh23.


[2 marks]

c.

0=2500π3-πh2        (M1)


Note:
Award (M1) for equating their derivative to zero. Follow through from part (c).


OR

sketch of dVdh        (M1)


Note:
 Award (M1) for a labelled sketch of dVdh with the curve/axes correctly labelled or the x-intercept explicitly indicated.


h=  28.9 cm  25003, 503, 5033, 28.8675       (A1)(ft)


Note: An unsupported 28.9 cm is awarded no marks. Graphing the function Vh is not an acceptable method and (M0)(A0) should be awarded. Follow through from part (c). Given the restraints of the question, h50 is not possible.


[2 marks]

d.

V= 2500×π×28.86753-π28.867533        (M1)

OR

13π40.8282×28.8675        (M1)


Note:
 Award (M1) for substituting their 28.8675 in the volume formula.


V=  50400 cm3  50383.3       (A1)(ft)(G2)


Note: Follow through from part (d).


[2 marks]

e.

S= π×2500-28.86752×50         (A1)(ft)(M1)


Note:
 Award (A1) for their correct radius seen 40.8248, 2500-28.86752.
Award (M1) for correctly substituted curved surface area formula for a cone.


S= 6410 cm2  6412.74       (A1)(ft)(G2)


Note: Follow through from parts (a) and (d).


[3 marks]

f.

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.

Syllabus sections

Topic 5 —Calculus » SL 5.8—Testing for max and min, optimisation. Points of inflexion
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Topic 5 —Calculus

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