| Exam Board | OCR |
|---|---|
| Module | PURE |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Topic | Stationary points and optimisation |
| Type | Show formula then optimise: cylinder/prism (single variable) |
| Difficulty | Standard +0.8 This is a multi-part optimization problem requiring geometric visualization, Pythagorean theorem to find slant heights, constraint equation manipulation, differentiation, and critical point analysis. While the individual techniques are standard A-level, the problem requires careful setup of the surface area constraint involving the slant height √(x²+1), algebraic manipulation to express V in the required form, and finding the specific form k=√2+1. The multiple steps and geometric reasoning place it moderately above average difficulty. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives6.02j Conservation with elastics: springs and strings |
| Answer | Marks | Guidance |
|---|---|---|
| \((V =)\frac{1}{2}x(2x)y = x^2y\) | B1 | 1.1 |
| Slant height of the roof is \(x\sqrt{2}\) | B1 | 3.1a |
| \((S =)2xy + 2\!\left(\tfrac{1}{2}(2x)x\right) + 2\!\left(yx\sqrt{2}\right)\) | M1\* | 2.1 |
| \(y = \dfrac{600-2x^2}{2x(1+\sqrt{2})} \Rightarrow V = x^2\!\left(\dfrac{300-x^2}{x(1+\sqrt{2})}\right)\) | M1dep\* | 3.3 |
| \(V = x(300-x^2)\!\left(\dfrac{(1-\sqrt{2})}{(1+\sqrt{2})(1-\sqrt{2})}\right)\) | M1 | 1.1 |
| \(V = x(300-x^2)\!\left(\dfrac{1-\sqrt{2}}{1-2}\right) = (\sqrt{2}-1)x(300-x^2)\) | A1 | 2.2a |
| Answer | Marks | Guidance |
|---|---|---|
| \(\dfrac{dV}{dx} = k(300-3x^2)\) | M1\* | 1.1 |
| \((k)(300-3x^2)=0 \Rightarrow x = \ldots\) | A1 | 1.1 |
| Sets \(\dfrac{dV}{dx}=0\) and attempts to find \(x\) | M1dep\* | 1.1 |
| \(x = 10\) cm | A1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(V = 828\ \text{cm}^3\) | B1 | 3.4 |
| Answer | Marks | Guidance |
|---|---|---|
| \(V\) (or \(y\)) must be positive or \(300 - x^2 > 0\) | M1 | 3.5b |
| so \(x\) cannot exceed \(\sqrt{300}\) cm | A1 | 1.1 |
## Question 7(a):
$(V =)\frac{1}{2}x(2x)y = x^2y$ | **B1** | 1.1 | Correct simplified expression for the volume
Slant height of the roof is $x\sqrt{2}$ | **B1** | 3.1a | Allow $\sqrt{2x^2}$
$(S =)2xy + 2\!\left(\tfrac{1}{2}(2x)x\right) + 2\!\left(yx\sqrt{2}\right)$ | **M1\*** | 2.1 | Attempt at surface area with at least three of the five faces correct – can be unsimplified
$y = \dfrac{600-2x^2}{2x(1+\sqrt{2})} \Rightarrow V = x^2\!\left(\dfrac{300-x^2}{x(1+\sqrt{2})}\right)$ | **M1dep\*** | 3.3 | Rearranges and makes $y$ the subject and substitutes to give an expression for $V$ in terms of $x$ only
$V = x(300-x^2)\!\left(\dfrac{(1-\sqrt{2})}{(1+\sqrt{2})(1-\sqrt{2})}\right)$ | **M1** | 1.1 | Rationalises the denominator correctly
$V = x(300-x^2)\!\left(\dfrac{1-\sqrt{2}}{1-2}\right) = (\sqrt{2}-1)x(300-x^2)$ | **A1** | 2.2a | $a=2,\ b=-1$
**[6 marks]**
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## Question 7(b):
$\dfrac{dV}{dx} = k(300-3x^2)$ | **M1\*** | 1.1 | **M1** for attempt at differentiation – both powers reduced by 1. Allow full marks ft their values of $a$ and $b$
$(k)(300-3x^2)=0 \Rightarrow x = \ldots$ | **A1** | 1.1 |
Sets $\dfrac{dV}{dx}=0$ and attempts to find $x$ | **M1dep\*** | 1.1 |
$x = 10$ cm | **A1** | 1.1 |
**[4 marks]**
---
## Question 7(c):
$V = 828\ \text{cm}^3$ | **B1** | 3.4 | cao. $828.4271247\ldots$
**[1 mark]**
---
## Question 7(d):
$V$ (or $y$) must be positive or $300 - x^2 > 0$ | **M1** | 3.5b | Explanation for constraint on $x$
so $x$ cannot exceed $\sqrt{300}$ cm | **A1** | 1.1 | Correct value; accept e.g. 17.3 or better
**[2 marks]**
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\includegraphics[max width=\textwidth, alt={}, center]{7fc02f90-8f8b-4153-bba1-dc0807124e96-5_421_944_251_242}
The diagram shows a model for the roof of a toy building. The roof is in the form of a solid triangular prism $A B C D E F$. The base $A C F D$ of the roof is a horizontal rectangle, and the crosssection $A B C$ of the roof is an isosceles triangle with $A B = B C$.
The lengths of $A C$ and $C F$ are $2 x \mathrm {~cm}$ and $y \mathrm {~cm}$ respectively, and the height of $B E$ above the base of the roof is $x \mathrm {~cm}$.
The total surface area of the five faces of the roof is $600 \mathrm {~cm} ^ { 2 }$ and the volume of the roof is $V \mathrm {~cm} ^ { 3 }$.
\begin{enumerate}[label=(\alph*)]
\item Show that $V = k x \left( 300 - x ^ { 2 } \right)$, where $k = \sqrt { a } + b$ and $a$ and $b$ are integers to be determined.
\item Use differentiation to determine the value of $x$ for which the volume of the roof is a maximum.
\item Find the maximum volume of the roof. Give your answer in $\mathrm { cm } ^ { 3 }$, correct to the nearest integer.
\item Explain why, for this roof, $x$ must be less than a certain value, which you should state.
\end{enumerate}
\hfill \mbox{\textit{OCR PURE Q7 [13]}}