Date | November 2020 | Marks available | 2 | 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 50 cm.
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.
Show that V=2500πh3-πh33.
Find dVdh.
Using your answer to part (c), find the value of h for which V is a maximum.
Find the maximum volume of the bird bath.
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.
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]
13×π×(2500-h2)×h OR 13×π×(√2500-h2)2×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]
(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]
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, 50√3, 50√33, 28.8675…) (A1)(ft)
Note: An unsupported 28.9 cm is awarded no marks. Graphing the function V(h) is not an acceptable method and (M0)(A0) should be awarded. Follow through from part (c). Given the restraints of the question, h≥50 is not possible.
[2 marks]
(V=) 2500×π×28.8675…3-π(28.8675…)33 (M1)
OR
13π(40.828…)2×28.8675… (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]
(S=) π×√2500-(28.8675…)2×50 (A1)(ft)(M1)
Note: Award (A1) for their correct radius seen (40.8248…, √2500-(28.8675…)2).
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]