| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2011 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Applied differentiation |
| Type | Optimise 3D shape dimensions |
| Difficulty | Standard +0.3 This is a standard C2 optimization problem with scaffolding. Part (a) guides students to derive the constraint equation using given volume, part (b) is routine differentiation and solving dL/dx=0, and part (c) asks for second derivative test. The algebra is straightforward and all steps are well-signposted, making this slightly easier than average. |
| Spec | 1.07n 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 |
|---|---|---|
| Working | Mark | Guidance |
| \(2x^2\!\left(\dfrac{L-12x}{4}\right)=81\) | B1 oe | |
| \(\Rightarrow x^2(L-12x)=162 \Rightarrow L = 12x + \dfrac{162}{x^2}\) | M1 | Rearranges their equation to make \(y\) the subject |
| A1 cso | Correct solution only. AG |
## Question 8(a) — Aliter Way 2:
| Working | Mark | Guidance |
|---|---|---|
| $2x^2\!\left(\dfrac{L-12x}{4}\right)=81$ | B1 oe | |
| $\Rightarrow x^2(L-12x)=162 \Rightarrow L = 12x + \dfrac{162}{x^2}$ | M1 | Rearranges their equation to make $y$ the subject |
| | A1 cso | Correct solution only. **AG** |
**[3 marks]**8.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{c9758792-ca4c-4837-bd7c-e695fe0c0cdf-12_662_719_127_609}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
A cuboid has a rectangular cross-section where the length of the rectangle is equal to twice its width, $x \mathrm {~cm}$, as shown in Figure 2.\\
The volume of the cuboid is 81 cubic centimetres.
\begin{enumerate}[label=(\alph*)]
\item Show that the total length, $L \mathrm {~cm}$, of the twelve edges of the cuboid is given by
$$L = 12 x + \frac { 162 } { x ^ { 2 } }$$
\item Use calculus to find the minimum value of $L$.
\item Justify, by further differentiation, that the value of $L$ that you have found is a minimum.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2011 Q8 [11]}}