Question 8 - A-Level Maths

Edexcel AS Paper 1 2018 June — Question 8 9 marks

Exam Board	Edexcel
Module	AS Paper 1 (AS Paper 1)
Year	2018
Session	June
Marks	9
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Stationary points and optimisation
Type	Optimise cost or profit model
Difficulty	Moderate -0.3 This is a standard optimization problem requiring differentiation to find a minimum, verification using the second derivative, and a brief modeling critique. The algebra is straightforward (equating dC/dv to zero gives a simple equation), and all steps follow a well-practiced routine for AS-level students. It's slightly easier than average because the differentiation is uncomplicated and the problem structure is highly familiar.
Spec	1.07n Stationary points: find maxima, minima using derivatives 1.07p Points of inflection: using second derivative

A lorry is driven between London and Newcastle.

In a simple model, the cost of the journey $\pounds C$ when the lorry is driven at a steady speed of $v$ kilometres per hour is $$C = \frac { 1500 } { v } + \frac { 2 v } { 11 } + 60$$

Find, according to this model,
1. the value of $v$ that minimises the cost of the journey,
2. the minimum cost of the journey.
  (Solutions based entirely on graphical or numerical methods are not acceptable.)
Prove by using $\frac { \mathrm { d } ^ { 2 } C } { \mathrm {~d} v ^ { 2 } }$ that the cost is minimised at the speed found in (a)(i).
State one limitation of this model.

Show mark scheme Show mark scheme source

Question 8:

Part (a)(i):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$C = \frac{1500}{v} + \frac{2v}{11} + 60 \Rightarrow \frac{dC}{dv} = -\frac{1500}{v^2} + \frac{2}{11}$	M1, A1	M1: Attempts to differentiate, dealing with powers of $v$ correctly. Look for $\frac{dC}{dv}$ in form $\frac{A}{v^2} + B$. Condone $\frac{dC}{dv}$ appearing as $\frac{dy}{dx}$ or not appearing at all.
Sets $\frac{dC}{dv} = 0 \Rightarrow v^2 = 8250$	M1	Sets $\frac{dC}{dv}=0$ (may be implied) and proceeds to equation of type $v^n = k, k>0$
$\Rightarrow v = \sqrt{8250} \Rightarrow v = 90.8 \text{ (km h}^{-1})$	A1	Or $5\sqrt{330}$ awrt $90.8$ km/h. As this is a speed, withhold mark for $v = \pm\sqrt{8250}$

Part (a)(ii):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
Substitutes $v = 90.8$ in $C = \frac{1500}{v} + \frac{2v}{11} + 60$	M1
Minimum cost $=$ awrt $(£)93$	A1 ft

Part (b):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
Finds $\frac{d^2C}{dv^2} = +\frac{3000}{v^3}$ at $v = 90.8$	M1
$\frac{d^2C}{dv^2} = (+0.004) > 0$ hence minimum (cost)	A1 ft

Part (c):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
It would be impossible to drive at this speed over the whole journey	B1

Question (a)(ii):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
Correct method of finding $C$ from their solution to $\frac{dC}{dv} = 0$	M1	Do not accept attempts using negative values of $v$. Award if $v = ..., C = ...$ where $v$ is their solution to (a)(i)
Minimum cost = awrt (£) 93	A1ft	Condone omission of units. Follow through on sensible values of $v$, $60 < v < 110$

Question (b) [Second derivative test]:

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
Finds $\frac{d^2C}{dv^2}$ following through on their $\frac{dC}{dv}$ and attempts to find its value/sign at their $v$	M1	Allow substitution of their answer to (a)(i) into $\frac{d^2C}{dv^2}$. Allow explanation of sign from its terms (as $v > 0$)
$\frac{d^2C}{dv^2} = +0.004 > 0$ hence minimum (cost). Alternatively $\frac{d^2C}{dv^2} = +\frac{3000}{v^3} > 0$ as $v > 0$	A1ft	Requires correct calculation/expression, correct statement and correct conclusion. Follow through on their $v$ $(v>0)$ and their $\frac{d^2C}{dv^2}$. Condone $\frac{d^2C}{dv^2}$ appearing as $\frac{d^2y}{dx^2}$ for M1 but for A1 correct notation must be used (accept $C''$)

Question (c):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
A limitation of the given model, e.g. it would be impossible to drive at this speed over the whole journey; traffic would mean you cannot drive at constant speed	B1	Any statement implying speed could not be constant is acceptable

# Question 8:

## Part (a)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $C = \frac{1500}{v} + \frac{2v}{11} + 60 \Rightarrow \frac{dC}{dv} = -\frac{1500}{v^2} + \frac{2}{11}$ | M1, A1 | M1: Attempts to differentiate, dealing with powers of $v$ correctly. Look for $\frac{dC}{dv}$ in form $\frac{A}{v^2} + B$. Condone $\frac{dC}{dv}$ appearing as $\frac{dy}{dx}$ or not appearing at all. |
| Sets $\frac{dC}{dv} = 0 \Rightarrow v^2 = 8250$ | M1 | Sets $\frac{dC}{dv}=0$ (may be implied) and proceeds to equation of type $v^n = k, k>0$ |
| $\Rightarrow v = \sqrt{8250} \Rightarrow v = 90.8 \text{ (km h}^{-1})$ | A1 | Or $5\sqrt{330}$ awrt $90.8$ km/h. As this is a speed, withhold mark for $v = \pm\sqrt{8250}$ |

## Part (a)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Substitutes $v = 90.8$ in $C = \frac{1500}{v} + \frac{2v}{11} + 60$ | M1 | |
| Minimum cost $=$ awrt $(£)93$ | A1 ft | |

## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Finds $\frac{d^2C}{dv^2} = +\frac{3000}{v^3}$ at $v = 90.8$ | M1 | |
| $\frac{d^2C}{dv^2} = (+0.004) > 0$ hence minimum (cost) | A1 ft | |

## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| It would be impossible to drive at this speed over the whole journey | B1 | |

## Question (a)(ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct method of finding $C$ from their solution to $\frac{dC}{dv} = 0$ | M1 | Do not accept attempts using negative values of $v$. Award if $v = ..., C = ...$ where $v$ is their solution to (a)(i) |
| Minimum cost = awrt (£) 93 | A1ft | Condone omission of units. Follow through on sensible values of $v$, $60 < v < 110$ |

---

## Question (b) [Second derivative test]:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Finds $\frac{d^2C}{dv^2}$ following through on their $\frac{dC}{dv}$ and attempts to find its value/sign at their $v$ | M1 | Allow substitution of their answer to (a)(i) into $\frac{d^2C}{dv^2}$. Allow explanation of sign from its terms (as $v > 0$) |
| $\frac{d^2C}{dv^2} = +0.004 > 0$ hence minimum (cost). Alternatively $\frac{d^2C}{dv^2} = +\frac{3000}{v^3} > 0$ as $v > 0$ | A1ft | Requires correct calculation/expression, correct statement and correct conclusion. Follow through on their $v$ $(v>0)$ and their $\frac{d^2C}{dv^2}$. Condone $\frac{d^2C}{dv^2}$ appearing as $\frac{d^2y}{dx^2}$ for M1 but for A1 correct notation must be used (accept $C''$) |

---

## Question (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| A limitation of the given model, e.g. it would be impossible to drive at this speed over the whole journey; traffic would mean you cannot drive at constant speed | B1 | Any statement implying speed could not be constant is acceptable |

---

Show LaTeX source

\begin{enumerate}
  \item A lorry is driven between London and Newcastle.
\end{enumerate}

In a simple model, the cost of the journey $\pounds C$ when the lorry is driven at a steady speed of $v$ kilometres per hour is

$$C = \frac { 1500 } { v } + \frac { 2 v } { 11 } + 60$$

(a) Find, according to this model,\\
(i) the value of $v$ that minimises the cost of the journey,\\
(ii) the minimum cost of the journey.\\
(Solutions based entirely on graphical or numerical methods are not acceptable.)\\
(b) Prove by using $\frac { \mathrm { d } ^ { 2 } C } { \mathrm {~d} v ^ { 2 } }$ that the cost is minimised at the speed found in (a)(i).\\
(c) State one limitation of this model.

\hfill \mbox{\textit{Edexcel AS Paper 1 2018 Q8 [9]}}

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