| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2015 |
| 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 optimization problem requiring surface area constraint, volume formula manipulation, differentiation of a polynomial, and second derivative test. All steps are routine C1 techniques with clear structure, making it slightly easier than average despite multiple parts. |
| Spec | 1.02z Models in context: use functions in modelling1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Surface area = curved surface + base: \(2\pi rh + \pi r^2 = 48\pi\) | M1 | Setting up correct surface area equation |
| \(h = \frac{48\pi - \pi r^2}{2\pi r}\) | A1 | Correct rearrangement |
| \(h = \frac{48 - r^2}{2r}\) | A1 | Simplified form |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(V = \pi r^2 h = \pi r^2 \times \frac{48 - r^2}{2r}\) | M1 | Substituting \(h\) into \(V = \pi r^2 h\) |
| \(V = \frac{\pi r(48 - r^2)}{2} = 24\pi r - \frac{\pi}{2}r^3\) | A1 | Correct completion with no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{dV}{dr} = 24\pi - \frac{3\pi}{2}r^2\) | M1 | Attempt to differentiate |
| \(\frac{dV}{dr} = 24\pi - \frac{3\pi}{2}r^2\) | A1 | Correct derivative |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Set \(\frac{dV}{dr} = 0\): \(24\pi - \frac{3\pi}{2}r^2 = 0\) | M1 | Setting derivative equal to zero |
| \(r^2 = 16\), so \(r = 4\) | A1 | Correct positive value |
| \(\frac{d^2V}{dr^2} = -3\pi r\) | M1 | Attempt second derivative or other valid method |
| At \(r=4\): \(\frac{d^2V}{dr^2} = -12\pi < 0\), therefore maximum | A1 | Correct conclusion with justification |
# Question 6:
## Part (a)(i):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Surface area = curved surface + base: $2\pi rh + \pi r^2 = 48\pi$ | M1 | Setting up correct surface area equation |
| $h = \frac{48\pi - \pi r^2}{2\pi r}$ | A1 | Correct rearrangement |
| $h = \frac{48 - r^2}{2r}$ | A1 | Simplified form |
## Part (a)(ii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $V = \pi r^2 h = \pi r^2 \times \frac{48 - r^2}{2r}$ | M1 | Substituting $h$ into $V = \pi r^2 h$ |
| $V = \frac{\pi r(48 - r^2)}{2} = 24\pi r - \frac{\pi}{2}r^3$ | A1 | Correct completion with no errors |
## Part (b)(i):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{dV}{dr} = 24\pi - \frac{3\pi}{2}r^2$ | M1 | Attempt to differentiate |
| $\frac{dV}{dr} = 24\pi - \frac{3\pi}{2}r^2$ | A1 | Correct derivative |
## Part (b)(ii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Set $\frac{dV}{dr} = 0$: $24\pi - \frac{3\pi}{2}r^2 = 0$ | M1 | Setting derivative equal to zero |
| $r^2 = 16$, so $r = 4$ | A1 | Correct positive value |
| $\frac{d^2V}{dr^2} = -3\pi r$ | M1 | Attempt second derivative or other valid method |
| At $r=4$: $\frac{d^2V}{dr^2} = -12\pi < 0$, therefore **maximum** | A1 | Correct conclusion with justification |
---6 The diagram shows a cylindrical container of radius $r \mathrm {~cm}$ and height $h \mathrm {~cm}$. The container has an open top and a circular base.\\
\includegraphics[max width=\textwidth, alt={}, center]{c7f38f7e-75aa-4b41-96fd-f38f968c225c-12_389_426_404_751}
The external surface area of the container's curved surface and base is $48 \pi \mathrm {~cm} ^ { 2 }$.\\
When the radius of the base is $r \mathrm {~cm}$, the volume of the container is $V \mathrm {~cm} ^ { 3 }$.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Find an expression for $h$ in terms of $r$.
\item Show that $V = 24 \pi r - \frac { \pi } { 2 } r ^ { 3 }$.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Find $\frac { \mathrm { d } V } { \mathrm {~d} r }$.
\item Find the positive value of $r$ for which $V$ is stationary, and determine whether this stationary value is a maximum value or a minimum value.\\[0pt]
[4 marks]
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C1 2015 Q6 [11]}}