AQA C1 2010 June — Question 6 11 marks

Exam BoardAQA
ModuleC1 (Core Mathematics 1)
Year2010
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicStationary points and optimisation
TypeShow formula then optimise: cylinder/prism (single variable)
DifficultyModerate -0.3 This is a structured optimization problem with extensive scaffolding through multiple parts. While it involves several steps (surface area constraint, volume formula, differentiation, second derivative test), each part is clearly signposted and uses standard C1 techniques. The algebraic manipulation is straightforward, and the 'show that' format removes problem-solving burden. Slightly easier than average due to heavy guidance, though the multi-step nature prevents it from being trivial.
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
6 The diagram shows a block of wood in the shape of a prism with triangular cross-section. The end faces are right-angled triangles with sides of lengths \(3 x \mathrm {~cm}\), \(4 x \mathrm {~cm}\) and \(5 x \mathrm {~cm}\), and the length of the prism is \(y \mathrm {~cm}\), as shown in the diagram. \includegraphics[max width=\textwidth, alt={}, center]{66813123-3876-4484-aad1-4bfc09bb1508-7_394_825_459_548} The total surface area of the five faces is \(144 \mathrm {~cm} ^ { 2 }\).
    1. Show that \(x y + x ^ { 2 } = 12\).
    2. Hence show that the volume of the block, \(V \mathrm {~cm} ^ { 3 }\), is given by $$V = 72 x - 6 x ^ { 3 }$$
    1. Find \(\frac { \mathrm { d } V } { \mathrm {~d} x }\).
    2. Show that \(V\) has a stationary value when \(x = 2\).
  3. Find \(\frac { \mathrm { d } ^ { 2 } V } { \mathrm {~d} x ^ { 2 } }\) and hence determine whether \(V\) has a maximum value or a minimum value when \(x = 2\).
    (2 marks)
Question 6:
Part (a)(i): Show that \(xy + x^2 = 12\)
AnswerMarks Guidance
WorkingMark Guidance
Area of two triangular faces: \(2 \times \frac{1}{2}(3x)(4x) = 12x^2\)M1 Correct area of triangular faces
Three rectangular faces: \(3xy + 4xy + 5xy = 12xy\)M1 Correct area of rectangular faces
\(12x^2 + 12xy = 144 \Rightarrow xy + x^2 = 12\)A1 Dividing through by 12, correct completion
Part (a)(ii): Show \(V = 72x - 6x^3\)
AnswerMarks Guidance
WorkingMark Guidance
\(V = \frac{1}{2}(3x)(4x)(y) = 6x^2 y\)M1 Correct volume expression
\(y = \frac{12 - x^2}{x}\), substituting: \(V = 6x^2 \cdot \frac{12-x^2}{x} = 72x - 6x^3\)A1 Correct completion
Part (b)(i): Find \(\frac{dV}{dx}\)
AnswerMarks Guidance
WorkingMark Guidance
\(\frac{dV}{dx} = 72 - 18x^2\)M1A1 M1 for attempting differentiation, A1 correct
Part (b)(ii): Show stationary value when \(x = 2\)
AnswerMarks Guidance
WorkingMark Guidance
\(72 - 18x^2 = 0 \Rightarrow x^2 = 4 \Rightarrow x = 2\)M1A1 Setting derivative to zero and solving
Part (c): Find \(\frac{d^2V}{dx^2}\) and determine nature
AnswerMarks Guidance
WorkingMark Guidance
\(\frac{d^2V}{dx^2} = -36x\)M1 Correct second derivative
When \(x=2\): \(\frac{d^2V}{dx^2} = -72 < 0\), therefore maximumA1 Correct conclusion with reason 
## Question 6:

**Part (a)(i):** Show that $xy + x^2 = 12$

| Working | Mark | Guidance |
|---------|------|----------|
| Area of two triangular faces: $2 \times \frac{1}{2}(3x)(4x) = 12x^2$ | M1 | Correct area of triangular faces |
| Three rectangular faces: $3xy + 4xy + 5xy = 12xy$ | M1 | Correct area of rectangular faces |
| $12x^2 + 12xy = 144 \Rightarrow xy + x^2 = 12$ | A1 | Dividing through by 12, correct completion |

**Part (a)(ii):** Show $V = 72x - 6x^3$

| Working | Mark | Guidance |
|---------|------|----------|
| $V = \frac{1}{2}(3x)(4x)(y) = 6x^2 y$ | M1 | Correct volume expression |
| $y = \frac{12 - x^2}{x}$, substituting: $V = 6x^2 \cdot \frac{12-x^2}{x} = 72x - 6x^3$ | A1 | Correct completion |

**Part (b)(i):** Find $\frac{dV}{dx}$

| Working | Mark | Guidance |
|---------|------|----------|
| $\frac{dV}{dx} = 72 - 18x^2$ | M1A1 | M1 for attempting differentiation, A1 correct |

**Part (b)(ii):** Show stationary value when $x = 2$

| Working | Mark | Guidance |
|---------|------|----------|
| $72 - 18x^2 = 0 \Rightarrow x^2 = 4 \Rightarrow x = 2$ | M1A1 | Setting derivative to zero and solving |

**Part (c):** Find $\frac{d^2V}{dx^2}$ and determine nature

| Working | Mark | Guidance |
|---------|------|----------|
| $\frac{d^2V}{dx^2} = -36x$ | M1 | Correct second derivative |
| When $x=2$: $\frac{d^2V}{dx^2} = -72 < 0$, therefore **maximum** | A1 | Correct conclusion with reason |

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6 The diagram shows a block of wood in the shape of a prism with triangular cross-section. The end faces are right-angled triangles with sides of lengths $3 x \mathrm {~cm}$, $4 x \mathrm {~cm}$ and $5 x \mathrm {~cm}$, and the length of the prism is $y \mathrm {~cm}$, as shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{66813123-3876-4484-aad1-4bfc09bb1508-7_394_825_459_548}

The total surface area of the five faces is $144 \mathrm {~cm} ^ { 2 }$.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Show that $x y + x ^ { 2 } = 12$.
\item Hence show that the volume of the block, $V \mathrm {~cm} ^ { 3 }$, is given by

$$V = 72 x - 6 x ^ { 3 }$$
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Find $\frac { \mathrm { d } V } { \mathrm {~d} x }$.
\item Show that $V$ has a stationary value when $x = 2$.
\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 $x = 2$.\\
(2 marks)
\end{enumerate}

\hfill \mbox{\textit{AQA C1 2010 Q6 [11]}}
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Q1 11 Q2 6 Q3 12 Q4 12 Q5 11 Q6 11 Q7 12