Date | November Example question | Marks available | 2 | Reference code | EXN.2.SL.TZ0.2 |
Level | Standard Level | Paper | Paper 2 | Time zone | Time zone 0 |
Command term | Hence or otherwise | Question number | 2 | Adapted from | N/A |
Question
A box of chocolates is to have a ribbon tied around it as shown in the diagram below.
The box is in the shape of a cuboid with a height of 3 cm. The length and width of the box are x and y cm.
After going around the box an extra 10 cm of ribbon is needed to form the bow.
The volume of the box is 450 cm3.
Find an expression for the total length of the ribbon L in terms of x and y.
Show that L=2x+300x+22
Find dLdx
Solve dLdx=0
Hence or otherwise find the minimum length of ribbon required.
Markscheme
* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.
L=2x+2y+12+10=2x+2y+22 A1A1
Note: A1 for 2x+2y and A1 for 12+10 or 22.
[2 marks]
V=3xy=450 A1
y=150x A1
L=2x+2(150x)+22 M1
L=2x+300x+22 AG
[3 marks]
L=2x+300x-1+22 (M1)
dLdx=2-300x2 A1A1
Note: A1 for 2 (and 0), A1 for 300x2.
[3 marks]
300x2=2 (M1)
x=√150=12.2 A1
[2 marks]
(M1)A1
[2 marks]