Date | May 2018 | Marks available | 4 | Reference code | 18M.1.sl.TZ2.9 |
Level | SL only | Paper | 1 | Time zone | TZ2 |
Command term | Show that | Question number | 9 | Adapted from | N/A |
Question
A closed cylindrical can with radius r centimetres and height h centimetres has a volume of 20\(\pi \) cm3.
The material for the base and top of the can costs 10 cents per cm2 and the material for the curved side costs 8 cents per cm2. The total cost of the material, in cents, is C.
Express h in terms of r.
Show that \(C = 20\pi {r^2} + \frac{{320\pi }}{r}\).
Given that there is a minimum value for C, find this minimum value in terms of \(\pi \).
Markscheme
correct equation for volume (A1)
eg \(\pi {r^2}h = 20\pi \)
\(h = \frac{{20}}{{{r^2}}}\) A1 N2
[2 marks]
attempt to find formula for cost of parts (M1)
eg 10 × two circles, 8 × curved side
correct expression for cost of two circles in terms of r (seen anywhere) A1
eg \(2\pi {r^2} \times 10\)
correct expression for cost of curved side (seen anywhere) (A1)
eg \(2\pi r \times h \times 8\)
correct expression for cost of curved side in terms of r A1
eg \(8 \times 2\pi r \times \frac{{20}}{{{r^2}}},\,\,\frac{{320\pi }}{{{r^2}}}\)
\(C = 20\pi {r^2} + \frac{{320\pi }}{r}\) AG N0
[4 marks]
recognize \(C' = 0\) at minimum (R1)
eg \(C' = 0,\,\,\frac{{{\text{d}}C}}{{{\text{d}}r}} = 0\)
correct differentiation (may be seen in equation)
\(C' = 40\pi r - \frac{{320\pi }}{{{r^2}}}\) A1A1
correct equation A1
eg \(40\pi r - \frac{{320\pi }}{{{r^2}}} = 0,\,\,40\pi r\frac{{320\pi }}{{{r^2}}}\)
correct working (A1)
eg \(40{r^3} = 320,\,\,{r^3} = 8\)
r = 2 (m) A1
attempt to substitute their value of r into C
eg \(20\pi \times 4 + 320 \times \frac{\pi }{2}\) (M1)
correct working
eg \(80\pi + 160\pi \) (A1)
\(240\pi \) (cents) A1 N3
Note: Do not accept 753.6, 753.98 or 754, even if 240\(\pi \) is seen.
[9 marks]