Date | November 2019 | Marks available | 2 | Reference code | 19N.2.SL.TZ0.T_6 |
Level | Standard Level | Paper | Paper 2 | Time zone | Time zone 0 |
Command term | Find | Question number | T_6 | Adapted from | N/A |
Question
The Maxwell Ohm Company is designing a portable Bluetooth speaker. The speaker is in the shape of a cylinder with a hemisphere at each end of the cylinder.
The dimensions of the speaker, in centimetres, are illustrated in the following diagram where r is the radius of the hemisphere, and l is the length of the cylinder, with r>0 and l≥0.
The Maxwell Ohm Company has decided that the speaker will have a surface area of 300 cm2.
The quality of sound from the speaker will improve as V increases.
Write down an expression for V, the volume (cm3) of the speaker, in terms of r, l and π.
Write down an equation for the surface area of the speaker in terms of r, l and π.
Given the design constraint that l=150-2πr2πr, show that V=150r-2πr33.
Find dVdr.
Using your answer to part (d), show that V is a maximum when r is equal to √75π cm.
Find the length of the cylinder for which V is a maximum.
Calculate the maximum value of V.
Use your answer to part (f) to identify the shape of the speaker with the best quality of sound.
Markscheme
(V=) 4πr33+πr2l (or equivalent) (A1)(A1)
Note: Award (A1) for either the volume of a hemisphere formula multiplied by 2 or the volume of a cylinder formula, and (A1) for completely correct expression. Accept equivalent expressions. Award at most (A1)(A0) if h is used instead of l.
[2 marks]
300=4πr2+2πrl (A1)(A1)(A1)
Note: Award (A1) for the surface area of a hemisphere multiplied by 2. Award (A1) for the surface area of a cylinder. Award (A1) for the addition of their formulas equated to 300. Award at most (A1)(A1)(A0) if h is used instead of l, unless already penalized in part (a).
[3 marks]
V=4πr33+πr2 (150-2πr2πr) (M1)
Note: Award (M1) for their correctly substituted formula for V.
V=4πr33+150r-2πr3 (M1)
Note: Award (M1) for correct expansion of brackets and simplification of the cylinder expression in V leading to the final answer.
V=150r-2πr33 (AG)
Note: The final line must be seen, with no incorrect working, for the second (M1) to be awarded.
[2 marks]
(dVdr=)150-2πr2 (A1)(A1)
Note: Award (A1) for 150. Award (A1) for -2πr2. Award maximum (A1)(A0) if extra terms seen.
[2 marks]
150-2πr2=0 OR dVdr=0 OR sketch of dVdrwith x-intercept indicated (M1)
Note: Award (M1) for equating their derivative to zero or a sketch of their derivative with x-intercept indicated.
r=√1502π OR r2=1502π (A1)
r=√75π (AG)
Note: The (AG) line must be seen for the preceding (A1) to be awarded.
[2 marks]
(l=)150-2π(√75π)2π(√75π) (M1)
Note: Award (M1) for correct substitution in the given formula for the length of the cylinder.
(l=) (A1)(G2)
Note: Award (M1)(A1) for correct substitution of the sf approximation leading to a correct answer of zero.
[2 marks]
OR (M1)
Note: Award (M1) for correct substitution in the formula for the volume of the speaker or the volume of a sphere.
(A1)(G2)
Note: Accept from use of sf value of . Award (M1)(A1)(ft) for correct substitution in their volume of speaker. Follow through from parts (a) and (f).
[2 marks]
sphere (spherical) (A1)(ft)
Note: Question requires the use of part (f) so if there is no answer to part (f), part (h) is awarded (A0). Follow through from their .
[1 mark]