Date | May 2022 | Marks available | 3 | Reference code | 22M.2.SL.TZ2.5 |
Level | Standard Level | Paper | Paper 2 | Time zone | Time zone 2 |
Command term | Sketch | Question number | 5 | Adapted from | N/A |
Question
A cafe makes x litres of coffee each morning. The cafe’s profit each morning, C, measured in dollars, is modelled by the following equation
C=x10(k2-3100x2)
where k is a positive constant.
The cafe’s manager knows that the cafe makes a profit of $426 when 20 litres of coffee are made in a morning.
The manager of the cafe wishes to serve as many customers as possible.
Find an expression for dCdx in terms of k and x.
Hence find the maximum value of C in terms of k. Give your answer in the form pk3, where p is a constant.
Find the value of k.
Use the model to find how much coffee the cafe should make each morning to maximize its profit.
Sketch the graph of C against x, labelling the maximum point and the x-intercepts with their coordinates.
Determine the maximum amount of coffee the cafe can make that will not result in a loss of money for the morning.
Markscheme
attempt to expand given expression (M1)
C=xk210-3x31000
dCdx=k210-9x21000 M1A1
Note: Award M1 for power rule correctly applied to at least one term and A1 for correct answer.
[3 marks]
equating their dCdx to zero (M1)
k210-9x21000=0
x2=100k29
x=10k3 (A1)
substituting their x back into given expression (M1)
Cmax=10k30(k2-300k2900)
Cmax=2k39 (0.222…k3) A1
[4 marks]
substituting 20 into given expression and equating to 426 M1
426=2010(k2-3100(20)2)
k=15 A1
[2 marks]
50 A1
[1 mark]
A1A1A1
Note: Award A1 for graph drawn for positive x indicating an increasing and then decreasing function, A1 for maximum labelled and A1 for graph passing through the origin and 86.6, marked on the x-axis or whose coordinates are given.
[3 marks]
setting their expression for C to zero OR choosing correct x-intercept on their graph of C (M1)
xmax=86.6 (86.6025…) litres A1
[2 marks]