Date | May 2019 | Marks available | 1 | Reference code | 19M.1.SL.TZ2.T_15 |
Level | Standard Level | Paper | Paper 1 (with calculator from previous syllabus) | Time zone | Time zone 2 |
Command term | Find | Question number | T_15 | Adapted from | N/A |
Question
A potter sells x vases per month.
His monthly profit in Australian dollars (AUD) can be modelled by
P(x)=−15x3+7x2−120,x⩾0.
Find the value of P if no vases are sold.
Differentiate P(x).
Hence, find the number of vases that will maximize the profit.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
−120 (AUD) (A1) (C1)
[1 mark]
−35x2+14x (A1)(A1) (C2)
Note: Award (A1) for each correct term. Award at most (A1)(A0) for extra terms seen.
[2 marks]
−35x2+14x=0 (M1)
Note: Award (M1) for equating their derivative to zero.
OR
sketch of their derivative (approximately correct shape) with x-intercept seen (M1)
2313(23.3,23.3333…,703) (A1)(ft)
Note: Award (C2) for 2313(23.3,23.3333…,703) seen without working.
23 (A1)(ft) (C3)
Note: Follow through from part (b).
[3 marks]