AQA C1 2012 June — Question 4 10 marks

Exam BoardAQA
ModuleC1 (Core Mathematics 1)
Year2012
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicStationary points and optimisation
TypeShow formula then optimise: cylinder/prism (single variable)
DifficultyModerate -0.3 This is a standard C1 optimization problem with heavily scaffolded steps: the constraint equation and volume formula are given to prove (not derive independently), differentiation is routine polynomial work, and the critical point value is provided to verify rather than find. While it requires multiple techniques (surface area, substitution, differentiation, second derivative test), each step is straightforward with no novel insight needed, making it slightly easier than average.
Spec1.02z Models in context: use functions in modelling1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives1.07p Points of inflection: using second derivative
4 The diagram shows a solid cuboid with sides of lengths \(x \mathrm {~cm} , 3 x \mathrm {~cm}\) and \(y \mathrm {~cm}\). \includegraphics[max width=\textwidth, alt={}, center]{dbc25177-4a28-480f-93d5-41acb2a2d28c-3_349_472_376_769} The total surface area of the cuboid is \(32 \mathrm {~cm} ^ { 2 }\).
    1. Show that \(3 x ^ { 2 } + 4 x y = 16\).
    2. Hence show that the volume, \(V \mathrm {~cm} ^ { 3 }\), of the cuboid is given by $$V = 12 x - \frac { 9 x ^ { 3 } } { 4 }$$
  2. Find \(\frac { \mathrm { d } V } { \mathrm {~d} x }\).
    1. Verify that a stationary value of \(V\) occurs when \(x = \frac { 4 } { 3 }\).
    2. Find \(\frac { \mathrm { d } ^ { 2 } V } { \mathrm {~d} x ^ { 2 } }\) and hence determine whether \(V\) has a maximum value or a minimum value when \(x = \frac { 4 } { 3 }\).
Question 4:
Part (a)(i):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(3x^2+3x^2+xy+xy+3xy+3xy\)M1 correct expression for surface area
\(6x^2+8xy=32\) \(\Rightarrow 3x^2+4xy=16\)A1 AG be convinced; \(2(3x^2+xy+3xy)=32\) etc
Part (a)(ii):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\((V=)3x^2y\) OEM1 correct volume in terms of \(x\) and \(y\)
\(=3x\left(\frac{16-3x^2}{4}\right)\) or \(=3x^2\left(\frac{16-3x^2}{4x}\right)\) OE
\(=12x-\frac{9x^3}{4}\)A1 CSO AG; be convinced that all working is correct
Part (b):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\frac{dV}{dx}=12-\frac{27}{4}x^2\)M1 A1 one of these terms correct; all correct with \(9\times3\) evaluated (no \(+c\) etc)
Part (c)(i):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(x=\frac{4}{3} \Rightarrow \frac{dV}{dx}=12-\frac{27}{4}\times\left(\frac{4}{3}\right)^2\)M1 attempt to sub \(x=\frac{4}{3}\) into 'their' \(\frac{dV}{dx}\)
\(\frac{dV}{dx}=12-\frac{27}{4}\times\frac{16}{9}=12-12\) or \(12-\frac{432}{36}=12-12\) or \(12-\frac{48}{4}=0\) etc
\(\frac{dV}{dx}=0 \Rightarrow\) stationary valueA1 CSO; shown \(=0\) plus statement
Part (c)(ii):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\frac{d^2V}{dx^2}=-\frac{27x}{2}\) OEB1\(\checkmark\) FT for 'their' \(\frac{dV}{dx}=a+bx^2\)
when \(x=\frac{4}{3}\), \(\frac{d^2V}{dx^2}<0 \Rightarrow\) maximumE1\(\checkmark\) or sub of \(x=\frac{4}{3}\) into 'their' \(\frac{d^2V}{dx^2} \Rightarrow\) maximum; E0 if numerical error seen
(FT *"minimum"* if their \(\frac{d^2V}{dx^2}>0\)) 
# Question 4:

## Part (a)(i):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $3x^2+3x^2+xy+xy+3xy+3xy$ | M1 | correct expression for surface area |
| $6x^2+8xy=32$ $\Rightarrow 3x^2+4xy=16$ | A1 | AG be convinced; $2(3x^2+xy+3xy)=32$ etc |

## Part (a)(ii):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $(V=)3x^2y$ OE | M1 | correct volume in terms of $x$ and $y$ |
| $=3x\left(\frac{16-3x^2}{4}\right)$ **or** $=3x^2\left(\frac{16-3x^2}{4x}\right)$ | | OE |
| $=12x-\frac{9x^3}{4}$ | A1 | CSO AG; be convinced that all working is correct |

## Part (b):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dV}{dx}=12-\frac{27}{4}x^2$ | M1 A1 | one of these terms correct; all correct with $9\times3$ evaluated (no $+c$ etc) |

## Part (c)(i):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $x=\frac{4}{3} \Rightarrow \frac{dV}{dx}=12-\frac{27}{4}\times\left(\frac{4}{3}\right)^2$ | M1 | attempt to sub $x=\frac{4}{3}$ into 'their' $\frac{dV}{dx}$ |
| $\frac{dV}{dx}=12-\frac{27}{4}\times\frac{16}{9}=12-12$ | | or $12-\frac{432}{36}=12-12$ or $12-\frac{48}{4}=0$ etc |
| $\frac{dV}{dx}=0 \Rightarrow$ stationary value | A1 | CSO; shown $=0$ plus statement |

## Part (c)(ii):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{d^2V}{dx^2}=-\frac{27x}{2}$ OE | B1$\checkmark$ | FT for 'their' $\frac{dV}{dx}=a+bx^2$ |
| when $x=\frac{4}{3}$, $\frac{d^2V}{dx^2}<0 \Rightarrow$ maximum | E1$\checkmark$ | **or** sub of $x=\frac{4}{3}$ into 'their' $\frac{d^2V}{dx^2} \Rightarrow$ maximum; E0 if numerical error seen |
| (FT *"minimum"* if their $\frac{d^2V}{dx^2}>0$) | | |
4 The diagram shows a solid cuboid with sides of lengths $x \mathrm {~cm} , 3 x \mathrm {~cm}$ and $y \mathrm {~cm}$.\\
\includegraphics[max width=\textwidth, alt={}, center]{dbc25177-4a28-480f-93d5-41acb2a2d28c-3_349_472_376_769}

The total surface area of the cuboid is $32 \mathrm {~cm} ^ { 2 }$.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Show that $3 x ^ { 2 } + 4 x y = 16$.
\item Hence show that the volume, $V \mathrm {~cm} ^ { 3 }$, of the cuboid is given by

$$V = 12 x - \frac { 9 x ^ { 3 } } { 4 }$$
\end{enumerate}\item Find $\frac { \mathrm { d } V } { \mathrm {~d} x }$.
\item \begin{enumerate}[label=(\roman*)]
\item Verify that a stationary value of $V$ occurs when $x = \frac { 4 } { 3 }$.
\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 $x = \frac { 4 } { 3 }$.
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{AQA C1 2012 Q4 [10]}}
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