| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2007 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Applied differentiation |
| Type | Optimise 3D shape dimensions |
| Difficulty | Moderate -0.5 This is a structured multi-part optimization problem with clear scaffolding through each step. While it requires setting up a surface area equation, substituting to eliminate variables, and using second derivatives, all steps are routine C1 techniques with no novel insight required. The 'show that' parts guide students through the algebra, making it slightly easier than average. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx1.08b Integrate x^n: where n != -1 and sums |
| Answer | Marks | Guidance |
|---|---|---|
| \(\Rightarrow x^2 + 3xh = 27\) | M1, A1 | Attempt at surface area (one slip); AG CSO |
| Answer | Marks | Guidance |
|---|---|---|
| \(h = \frac{27 - x^2}{3x}\) or \(h = \frac{9 - x}{x} - \frac{3}{}\) etc | B1 | Any correct form |
| Answer | Marks | Guidance |
|---|---|---|
| \(V = 2x^2h = 18x - \frac{2x^3}{3}\) | B1 | AG (watch fudging) condone omission of brackets |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dV}{dx} = 18 - 2x^2\) | M1, A1 | One term correct "their" \(V\); All correct unsimplified \(18 - 6x^2/3\) |
| Answer | Marks | Guidance |
|---|---|---|
| Shown to equal \(0\) plus statement that this implies a stationary point if verifying | M1, A1 | Or attempt to solve their \(\frac{dV}{dx} = 0\); CSO Condone \(x = \pm 3\) or \(x = 3\) if solving |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{d^2V}{dx^2} = -4x\) \((= -12)\) | B1✓ | FT their \(\frac{dV}{dx}\) |
| \(\frac{d^2V}{dx^2} < 0\) at stationary point \(\Rightarrow\) maximum | E1✓ | 2 marks |
## 5(a)(i)
$2x^2 + 2xh + 4xh (= 54)$
$\Rightarrow x^2 + 3xh = 27$ | M1, A1 | Attempt at surface area (one slip); AG CSO
## 5(a)(ii)
$h = \frac{27 - x^2}{3x}$ or $h = \frac{9 - x}{x} - \frac{3}{}$ etc | B1 | Any correct form
## 5(a)(iii)
$V = 2x^2h = 18x - \frac{2x^3}{3}$ | B1 | AG (watch fudging) condone omission of brackets
## 5(b)(i)
$\frac{dV}{dx} = 18 - 2x^2$ | M1, A1 | One term correct "their" $V$; All correct unsimplified $18 - 6x^2/3$
## 5(b)(ii)
Sub $x = 3$ into their $\frac{dV}{dx}$
Shown to equal $0$ plus **statement that this implies a stationary point** if verifying | M1, A1 | Or attempt to solve their $\frac{dV}{dx} = 0$; CSO Condone $x = \pm 3$ or $x = 3$ if solving
## 5(c)
$\frac{d^2V}{dx^2} = -4x$ $(= -12)$ | B1✓ | FT their $\frac{dV}{dx}$
$\frac{d^2V}{dx^2} < 0$ at stationary point $\Rightarrow$ maximum | E1✓ | 2 marks | FT their second derivative conclusion; If "their" $\frac{d^2V}{dx^2} > 0 \Rightarrow$ minimum etc
**Total: 10**
---5 The diagram shows an open-topped water tank with a horizontal rectangular base and four vertical faces. The base has width $x$ metres and length $2 x$ metres, and the height of the tank is $h$ metres.\\
\includegraphics[max width=\textwidth, alt={}, center]{33da89e2-f74f-4d5a-8bbd-ceaa728b6c34-4_403_410_477_792}
The combined internal surface area of the base and four vertical faces is $54 \mathrm {~m} ^ { 2 }$.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Show that $x ^ { 2 } + 3 x h = 27$.
\item Hence express $h$ in terms of $x$.
\item Hence show that the volume of water, $V \mathrm {~m} ^ { 3 }$, that the tank can hold when full is given by
$$V = 18 x - \frac { 2 x ^ { 3 } } { 3 }$$
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Find $\frac { \mathrm { d } V } { \mathrm {~d} x }$.
\item Verify that $V$ has a stationary value when $x = 3$.
\end{enumerate}\item Find $\frac { \mathrm { d } ^ { 2 } V } { \mathrm {~d} x ^ { 2 } }$ and hence determine whether $V$ has a maximum value or a minimum value when $x = 3$.\\
(2 marks)
\end{enumerate}
\hfill \mbox{\textit{AQA C1 2007 Q5 [10]}}