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Date	May 2021	Marks available	5	Reference code	21M.1.SL.TZ2.13
Level	Standard Level	Paper	Paper 1	Time zone	Time zone 2
Command term	Find	Question number	13	Adapted from	N/A

Question

A company produces and sells electric cars. The company’s profit, $P$ , in thousands of dollars, changes based on the number of cars, $x$ , they produce per month.

The rate of change of their profit from producing $x$ electric cars is modelled by

$\frac{d P}{d x} = - 1.6 x + 48, x \geq 0$ .

The company makes a profit of $260$ (thousand dollars) when they produce $15$ electric cars.

Find an expression for $P$ in terms of $x$ .

[5]

The company regularly increases the number of cars it produces.

Describe how their profit changes if they increase production to over $30$ cars per month and up to $50$ cars per month. Justify your answer.

[2]

Markscheme

recognition of need to integrate (eg reverse power rule or integral symbol) (M1)

$P (x) = - 0.8 x^{2} + 48 x (+ c)$ A1A1

$260 = - 0.8 \times {(15)}^{2} + 48 \times (15) + c$ (M1)

Note: Award M1 for correct substitution of $x = 15$ and $P = 260$ . A constant of integration must be seen (can be implied by a correct answer).

$c = - 280$

$P (x) = - 0.8 x^{2} + 48 x - 280$ A1

[5 marks]

profit will decrease (with each new car produced) A1

EITHER

because the profit function is decreasing / the gradient is negative / the rate of change of $P$ is negative R1

$\int_{30}^{50} - 1.6 x + 48 (d x) = - 320$ R1

evidence of finding $P (30) = 440$ and $P (50) = 120$ R1

Note: Award at most R1A0 if $P (30)$ or $P (50)$ or both have incorrect values.

[2 marks]

Examiners report

[N/A]

Syllabus sections

Topic 5—Calculus » SL 5.5—Integration introduction, areas between curve and x axis

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