Date | May 2018 | Marks available | 9 | Reference code | 18M.1.sl.TZ2.9 |
Level | SL only | Paper | 1 | Time zone | TZ2 |
Command term | Find | Question number | 9 | Adapted from | N/A |
Question
A closed cylindrical can with radius r centimetres and height h centimetres has a volume of 20ππ 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=20πr2+320πrC=20πr2+320πr.
Given that there is a minimum value for C, find this minimum value in terms of ππ.
Markscheme
correct equation for volume (A1)
eg πr2h=20ππr2h=20π
h=20r2h=20r2 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 2πr2×102πr2×10
correct expression for cost of curved side (seen anywhere) (A1)
eg 2πr×h×82πr×h×8
correct expression for cost of curved side in terms of r A1
eg 8×2πr×20r2,320πr28×2πr×20r2,320πr2
C=20πr2+320πrC=20πr2+320πr AG N0
[4 marks]
recognize C′=0 at minimum (R1)
eg C′=0,dCdr=0
correct differentiation (may be seen in equation)
C′=40πr−320πr2 A1A1
correct equation A1
eg 40πr−320πr2=0,40πr320πr2
correct working (A1)
eg 40r3=320,r3=8
r = 2 (m) A1
attempt to substitute their value of r into C
eg 20π×4+320×π2 (M1)
correct working
eg 80π+160π (A1)
240π (cents) A1 N3
Note: Do not accept 753.6, 753.98 or 754, even if 240π is seen.
[9 marks]