Edexcel C2 — Question 38 10 marks

Exam BoardEdexcel
ModuleC2 (Core Mathematics 2)
Marks10
PaperDownload PDF ↗
TopicTangents, normals and gradients
TypeOptimization with constraints
DifficultyModerate -0.3 This is a straightforward optimization problem requiring differentiation of a simple function (power and reciprocal terms), solving a cubic equation that factors nicely, and applying the second derivative test. While it involves multiple steps across three parts, each technique is standard C2 material with no novel insight required, making it slightly easier than average.
Spec1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives

On a journey, the average speed of a car is \(v\) m s\(^{-1}\). For \(v \geq 5\), the cost per kilometre, \(C\) pence, of the journey is modelled by $$C = \frac{160}{v} + \frac{v^2}{100}.$$ Using this model,
  1. show, by calculus, that there is a value of \(v\) for which \(C\) has a stationary value, and find this value of \(v\). [5]
  2. Justify that this value of \(v\) gives a minimum value of \(C\). [2]
  3. Find the minimum value of \(C\) and hence find the minimum cost of a 250 km car journey. [3]

On a journey, the average speed of a car is $v$ m s$^{-1}$. For $v \geq 5$, the cost per kilometre, $C$ pence, of the journey is modelled by
$$C = \frac{160}{v} + \frac{v^2}{100}.$$

Using this model,

\begin{enumerate}[label=(\alph*)]
\item show, by calculus, that there is a value of $v$ for which $C$ has a stationary value, and find this value of $v$. [5]

\item Justify that this value of $v$ gives a minimum value of $C$. [2]

\item Find the minimum value of $C$ and hence find the minimum cost of a 250 km car journey. [3]
\end{enumerate}

\hfill \mbox{\textit{Edexcel C2  Q38 [10]}}