| Exam Board | OCR |
|---|---|
| Module | H240/02 (Pure Mathematics and Statistics) |
| Year | 2019 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Topic | Stationary points and optimisation |
| Type | Optimise geometric shape surface area/volume |
| Difficulty | Standard +0.8 This is a constrained optimization problem requiring multiple steps: using Pythagoras to relate r, h, and l; substituting the surface area constraint; algebraic manipulation to express V in terms of r alone; then differentiation and solving a non-trivial equation involving r^4. The algebra is moderately demanding and the optimization requires careful handling of the constraint, placing it above average difficulty but not at the level requiring novel geometric insight. |
| Spec | 1.02z Models in context: use functions in modelling1.07b Gradient as rate of change: dy/dx notation1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(l = \frac{4}{r}\) or \(l = \frac{4\pi}{r\pi}\) exactly (not \(lr = 4\)) | B1 | Express \(l\) correctly in terms of \(r\); May be implied; Allow \(l = \frac{4-r^2}{r}\) oe B1 |
| \(h = \sqrt{\frac{16}{r^2} - r^2}\) or \(\frac{\sqrt{16-r^4}}{r}\) oe | B1 | Express \(h\) (or \(h^2\)) correctly in terms of \(r\) alone; or \(h^2 = \frac{16}{r^2} - r^2\) or \(\frac{16-r^4}{r^2}\) |
| \(V = \frac{1}{3}\pi r^2\sqrt{\frac{16}{r^2}-r^2}\) or \(\frac{1}{3}\pi r^2\frac{\sqrt{16-r^4}}{r}\) oe | M1 | Sub their \(h\) (in terms of \(r\) alone) into \(\frac{1}{3}\pi r^2 h\); Must see a correct previous expression in terms of \(r\) only, and the answer |
| \(\left(= \frac{\pi}{3}\sqrt{16r^2 - r^6}\right)\) AG | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{d}{dr}\left(\frac{\pi}{3}\sqrt{16r^2 - r^6}\right)\) | M1 | Attempt differentiate \(V\) or \(\frac{V}{\pi}\) or \(3V\); or \(\frac{3V}{\pi}\) or \(\sqrt{16r^2-r^6}\) or \(16r^2-r^6\) |
| \(\frac{\pi(32r - 6r^5)}{3\times2\sqrt{16r^2-r^6}} = 0\) oe (Their derivative \(= 0\)) | A1 | Correct derivative of one of the above; Condone missing brackets; All subsequent marks can be scored even if this A1 not scored |
| \(r = \frac{2}{\sqrt[4]{3}}\) or \(\sqrt[4]{\frac{16}{3}}\) oe or \(1.52\) (3 sf). Allow 1.5; or \(r^2 = \frac{4}{\sqrt{3}}\) | A1 | Lose this mark if incorrect values of \(r\) also given, eg \(r = \pm2\) obtained from \((16r^2-r^6)^{-\frac{1}{2}} = 0\); Allow without \(r=0\) |
| \(r = -\frac{2}{\sqrt[4]{3}}\) or \(-1.52\) invalid OR \(r = 0\) invalid or \(r > 0\) | B1f | Comment needed, about their negative \(r\) (ft) or about \(r = 0\) |
| \(V_{\max} = \frac{\pi}{3}\sqrt{16\times1.51967^2 - 1.51967^6}\) | T & I: 5.20 (3sf) SC B2; 5.2 (2 sf) SC B1 | |
| Max \(V = 5.20\) (3 sf). Allow 5.2 or a.r.t. 5.2 | A1 | Condone \(V = 5.20\) m\(^3\) |
# Question 5:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $l = \frac{4}{r}$ or $l = \frac{4\pi}{r\pi}$ exactly (not $lr = 4$) | B1 | Express $l$ correctly in terms of $r$; May be implied; Allow $l = \frac{4-r^2}{r}$ oe B1 |
| $h = \sqrt{\frac{16}{r^2} - r^2}$ or $\frac{\sqrt{16-r^4}}{r}$ oe | B1 | Express $h$ (or $h^2$) correctly in terms of $r$ alone; or $h^2 = \frac{16}{r^2} - r^2$ or $\frac{16-r^4}{r^2}$ |
| $V = \frac{1}{3}\pi r^2\sqrt{\frac{16}{r^2}-r^2}$ or $\frac{1}{3}\pi r^2\frac{\sqrt{16-r^4}}{r}$ oe | M1 | Sub their $h$ (in terms of $r$ alone) into $\frac{1}{3}\pi r^2 h$; Must see a correct previous expression in terms of $r$ only, and the answer |
| $\left(= \frac{\pi}{3}\sqrt{16r^2 - r^6}\right)$ **AG** | A1 | |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{d}{dr}\left(\frac{\pi}{3}\sqrt{16r^2 - r^6}\right)$ | M1 | Attempt differentiate $V$ or $\frac{V}{\pi}$ or $3V$; or $\frac{3V}{\pi}$ or $\sqrt{16r^2-r^6}$ or $16r^2-r^6$ |
| $\frac{\pi(32r - 6r^5)}{3\times2\sqrt{16r^2-r^6}} = 0$ oe (Their derivative $= 0$) | A1 | Correct derivative of one of the above; Condone missing brackets; All subsequent marks can be scored even if this A1 not scored |
| $r = \frac{2}{\sqrt[4]{3}}$ or $\sqrt[4]{\frac{16}{3}}$ oe or $1.52$ (3 sf). Allow 1.5; or $r^2 = \frac{4}{\sqrt{3}}$ | A1 | Lose this mark if incorrect values of $r$ also given, eg $r = \pm2$ obtained from $(16r^2-r^6)^{-\frac{1}{2}} = 0$; Allow without $r=0$ |
| $r = -\frac{2}{\sqrt[4]{3}}$ or $-1.52$ invalid OR $r = 0$ invalid or $r > 0$ | B1f | Comment needed, about their negative $r$ (ft) or about $r = 0$ |
| $V_{\max} = \frac{\pi}{3}\sqrt{16\times1.51967^2 - 1.51967^6}$ | | T & I: 5.20 (3sf) SC B2; 5.2 (2 sf) SC B1 |
| Max $V = 5.20$ (3 sf). Allow 5.2 or a.r.t. 5.2 | A1 | Condone $V = 5.20$ m$^3$ |
---5\\
\includegraphics[max width=\textwidth, alt={}, center]{1a0e0afb-81be-45d1-8c86-f98e508e9a49-06_371_339_255_251}
For a cone with base radius $r$, height $h$ and slant height $l$, the following formulae are given.\\
Curved surface area, $S = \pi r l$\\
Volume, $V = \frac { 1 } { 3 } \pi r ^ { 2 } h$\\
A container is to be designed in the shape of an inverted cone with no lid. The base radius is $r \mathrm {~m}$ and the volume is $V \mathrm {~m} ^ { 3 }$.
The area of the material to be used for the cone is $4 \pi \mathrm {~m} ^ { 2 }$.
\begin{enumerate}[label=(\alph*)]
\item Show that $V = \frac { 1 } { 3 } \pi \sqrt { 16 r ^ { 2 } - r ^ { 6 } }$.
\item In this question you must show detailed reasoning.
It is given that $V$ has a maximum value for a certain value of $r$.\\
Find the maximum value of $V$, giving your answer correct to 3 significant figures.
\end{enumerate}
\hfill \mbox{\textit{OCR H240/02 2019 Q5 [9]}}