| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Optimization with constraints |
| Difficulty | Standard +0.3 This is a standard C2 optimization problem with straightforward steps: forming an equation from volume, substituting to get surface area, differentiating, and using the second derivative test. All techniques are routine for this module, though it requires careful algebraic manipulation across multiple parts. Slightly easier than average due to its predictable structure and clear signposting through parts (a)-(e). |
| 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/dx1.07p Points of inflection: using second derivative |
| Answer | Marks | Guidance |
|---|---|---|
| \(2x^2h = 1030\), \(h = \frac{515}{x^2}\) | M1, A1 | (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(A = 4x^2 + 6xh\) (or unsimplified version) | B1 | |
| \(A = 4x^2 + \frac{3090}{x}\) | M1 A1 | (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dA}{dx} = 8x - 3090x^{-2}\) | M1 A1 | |
| \(8x - 3090x^{-2} = 0\) | M1 | |
| \(x^3 = (386.25)\) | M1 | |
| \(x = 7.283\) (7.28, 7.3) | A1 | (5 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(A = 4 \times 7.28^2 + \frac{3090}{7.28}\) | M1 | |
| \(= 636.4\) cm² | A1 | (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Second derivative \(= 8 + 6180x^{-3}\) | M1 | |
| Correct deriv. \(> 0\), \(\therefore\) Min | A1 | (2 marks) |
## (a)
$2x^2h = 1030$, $h = \frac{515}{x^2}$ | M1, A1 | (2 marks)
## (b)
$A = 4x^2 + 6xh$ (or unsimplified version) | B1 |
$A = 4x^2 + \frac{3090}{x}$ | M1 A1 | (3 marks)
## (c)
$\frac{dA}{dx} = 8x - 3090x^{-2}$ | M1 A1 |
$8x - 3090x^{-2} = 0$ | M1 |
$x^3 = (386.25)$ | M1 |
$x = 7.283$ (7.28, 7.3) | A1 | (5 marks)
## (d)
$A = 4 \times 7.28^2 + \frac{3090}{7.28}$ | M1 |
$= 636.4$ cm² | A1 | (2 marks)
## (e)
Second derivative $= 8 + 6180x^{-3}$ | M1 |
Correct deriv. $> 0$, $\therefore$ Min | A1 | (2 marks)
**Total: (14 marks)**\includegraphics{figure_3}
A manufacturer produces cartons for fruit juice. Each carton is in the shape of a closed cuboid with base dimensions $2x$ cm by $x$ cm and height $h$ cm, as shown in Fig. 3.
Given that the capacity of a carton has to be 1030 cm$^3$,
\begin{enumerate}[label=(\alph*)]
\item express $h$ in terms of $x$, [2]
\item show that the surface area, $A$ cm$^2$, of a carton is given by $A = 4x^2 + \frac{3090}{x}$. [3]
\end{enumerate}
The manufacturer needs to minimise the surface area of a carton.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Use calculus to find the value of $x$ for which $A$ is a minimum. [5]
\item Calculate the minimum value of $A$. [2]
\item Prove that this value of $A$ is a minimum. [2]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 Q8 [14]}}