| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2007 |
| Session | January |
| 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 straightforward C2 optimization question with a simple rational function. Students differentiate, set equal to zero, solve a basic equation (v² = 4900), verify with second derivative, and substitute back. All steps are routine textbook procedures with no conceptual challenges or novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02z Models in context: use functions in modelling1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dC}{dv} = -1400v^{-2} + \frac{2}{7}\) | M1, A1 | Attempt to differentiate \(v^n \to v^{n-1}\); must be differentiating function of form \(av^{-1}+bv\). Note: \((-1400v^{-2}+\frac{2}{7}+c\) is A0) |
| \(-1400v^{-2} + \frac{2}{7} = 0\) | M1 | Their \(\frac{dC}{dv}=0\); can be implied by \(\frac{dC}{dv}=P+Q \to P=\pm Q\) |
| \(v^2 = 4900\) | dM1 | Dependent on both previous M marks; attempt to rearrange into form \(v^n =\) number or \(v^n - \text{number}=0\), \(n\neq 0\) |
| \(v = 70\) | A1cso | \(v=70\) cso, but allow \(v=\pm 70\); \(v=70\) km per h also acceptable. Answer only is 0/5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{d^2C}{dv^2} = 2800v^{-3}\) | M1 | Attempt to differentiate \(\frac{dC}{dv}\); \(v^n \to v^{n-1}\) (including \(v^0 \to 0\)) |
| \(v=70\), \(\frac{d^2C}{dv^2} > 0 \Rightarrow\) minimum | A1ft | \(\frac{d^2C}{dv^2}\) must be correct; ft only from their value of \(v\) provided \(v\) is positive. Must have some indication their value of \(v\) is being used. Statement "\(\frac{d^2C}{dv^2}>0\)" sufficient provided \(2800v^{-3}>0\) for their \(v\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(v=70\), \(C = \frac{1400}{70} + \frac{2\times70}{7}\) | M1 | Substitute their value of \(v\) thought to give \(C_{\min}\) (independent of method and part) |
| \(C = 40\) | A1 | 40 or £40; must have part (a) completely correct (all 5 marks) to gain this A1. Answer only gains M1A1 provided part (a) completely correct |
## Question 8:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dC}{dv} = -1400v^{-2} + \frac{2}{7}$ | M1, A1 | Attempt to differentiate $v^n \to v^{n-1}$; must be differentiating function of form $av^{-1}+bv$. Note: $(-1400v^{-2}+\frac{2}{7}+c$ is A0) |
| $-1400v^{-2} + \frac{2}{7} = 0$ | M1 | Their $\frac{dC}{dv}=0$; can be implied by $\frac{dC}{dv}=P+Q \to P=\pm Q$ |
| $v^2 = 4900$ | dM1 | Dependent on both previous M marks; attempt to rearrange into form $v^n =$ number or $v^n - \text{number}=0$, $n\neq 0$ |
| $v = 70$ | A1cso | $v=70$ cso, but allow $v=\pm 70$; $v=70$ km per h also acceptable. Answer only is 0/5 |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{d^2C}{dv^2} = 2800v^{-3}$ | M1 | Attempt to differentiate $\frac{dC}{dv}$; $v^n \to v^{n-1}$ (including $v^0 \to 0$) |
| $v=70$, $\frac{d^2C}{dv^2} > 0 \Rightarrow$ minimum | A1ft | $\frac{d^2C}{dv^2}$ must be correct; ft only from their value of $v$ provided $v$ is positive. Must have some indication their value of $v$ is being used. Statement "$\frac{d^2C}{dv^2}>0$" sufficient provided $2800v^{-3}>0$ for their $v$ |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $v=70$, $C = \frac{1400}{70} + \frac{2\times70}{7}$ | M1 | Substitute their value of $v$ thought to give $C_{\min}$ (independent of method and part) |
| $C = 40$ | A1 | 40 or £40; must have part (a) completely correct (all 5 marks) to gain this A1. Answer only gains M1A1 provided part (a) completely correct |
---\begin{enumerate}
\item A diesel lorry is driven from Birmingham to Bury at a steady speed of v kilometres per hour. The total cost of the journey, $\pounds C$, is given by
\end{enumerate}
$$C = \frac { 1400 } { v } + \frac { 2 v } { 7 } .$$
(a) Find the value of $v$ for which $C$ is a minimum.\\
(b) Find $\frac { \mathrm { d } ^ { 2 } C } { \mathrm {~d} v ^ { 2 } }$ and hence verify that $C$ is a minimum for this value of $v$.\\
(c) Calculate the minimum total cost of the journey.\\
\hfill \mbox{\textit{Edexcel C2 2007 Q8 [9]}}