| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2019 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Stationary points and optimisation |
| Type | Show formula then optimise: composite/irregular shape |
| Difficulty | Standard +0.3 This is a standard optimization problem with a volume constraint requiring routine calculus techniques. Part (a) involves algebraic manipulation to derive the surface area formula (shown), part (b) requires differentiating and solving dA/dr = 0, and part (c) is simple substitution. While multi-step, it follows a well-practiced template with no novel insights required, making it slightly easier than average. |
| Spec | 1.02z Models in context: use functions in modelling1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| States or uses \(6=\pi r^2 h+\dfrac{2}{3}\pi r^3\) | B1 | 1.1a |
| \(A=2\pi r^2+2\pi r\!\left(\dfrac{6}{\pi r^2}-\dfrac{2}{3}r\right)+\pi r^2\) | M1, A1 | 3.1a, 1.1b — Complete substitution of \(h=...\) etc. into \(A=\lambda\pi r^2+\mu\pi rh\) |
| \(A=3\pi r^2+\dfrac{12}{r}-\dfrac{4}{3}\pi r^2\Rightarrow A=\dfrac{12}{r}+\dfrac{5}{3}\pi r^2\) * | A1* | 2.1 — Proceeds rigorously to given result |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\left\{A=12r^{-1}+\dfrac{5}{3}\pi r^2\Rightarrow\right\}\ \dfrac{\mathrm{d}A}{\mathrm{d}r}=-12r^{-2}+\dfrac{10}{3}\pi r\) | M1, A1 | 3.4, 1.1b — Differentiates \(\frac{\lambda}{r}+\mu r^2\) to give \(\alpha r^{-2}+\beta r\) |
| \(\left\{\dfrac{\mathrm{d}A}{\mathrm{d}r}=0\Rightarrow\right\}\ -\dfrac{12}{r^2}+\dfrac{10}{3}\pi r=0\Rightarrow-36+10\pi r^3=0\Rightarrow r^{\pm3}=...\left\{=\dfrac{18}{5\pi}\right\}\) | M1 | 2.1 — Sets \(\dfrac{\mathrm{d}A}{\mathrm{d}r}=0\) and rearranges to \(r^{\pm3}=k\), \(k\neq0\) |
| \(r=1.046447736...\Rightarrow r=1.05\ \text{(m)}\) (3 sf) or awrt 1.05 (m) | A1 | 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(A_{\min}=\dfrac{12}{(1.046...)}+\dfrac{5}{3}\pi(1.046...)^2\) | M1 | 3.4 |
| \(\{A_{\min}=17.20...\Rightarrow\}\ A=17\ (\text{m}^2)\) or \(A=\) awrt \(17\ (\text{m}^2)\) | A1ft | 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Substitutes \(r = 1.046...\) (found from solving \(\frac{dA}{dr} = 0\) in part (b)) into \(A = \frac{12}{r} + \frac{5}{3}\pi r^2\) | M1 | Give M0 for substituting \(r\) found from \(\frac{d^2A}{dr^2} = 0\) or using \(\frac{d^2A}{dr^2}\) into the model |
| \(\{A =\}\ 17\) or \(\{A =\}\) awrt 17 (ignoring units) | A1ft | Valid only for \(0.6 \leq r \leq 1.3\) where \(r\) found from \(\frac{dA}{dr} = 0\); Give M1A1 for \(A = 17\ \text{m}^2\) or \(A =\) awrt \(17\ \text{m}^2\) from no working |
# Question 13:
## Part (a)
| Working | Mark | Guidance |
|---------|------|----------|
| States or uses $6=\pi r^2 h+\dfrac{2}{3}\pi r^3$ | B1 | 1.1a |
| $A=2\pi r^2+2\pi r\!\left(\dfrac{6}{\pi r^2}-\dfrac{2}{3}r\right)+\pi r^2$ | M1, A1 | 3.1a, 1.1b — Complete substitution of $h=...$ etc. into $A=\lambda\pi r^2+\mu\pi rh$ |
| $A=3\pi r^2+\dfrac{12}{r}-\dfrac{4}{3}\pi r^2\Rightarrow A=\dfrac{12}{r}+\dfrac{5}{3}\pi r^2$ * | A1* | 2.1 — Proceeds rigorously to given result |
**Notes for (a):**
- B1: States or uses volume formula
- M1: Complete process of substituting their $h=...$ or $\pi h=...$ or $\pi rh=...$ or $rh=...$ into $A=\lambda\pi r^2+\mu\pi rh$, $\lambda,\mu\neq0$
- A1: Correct simplified or unsimplified $\{A=\}\ 2\pi r^2+2\pi r\!\left(\dfrac{6}{\pi r^2}-\dfrac{2}{3}r\right)+\pi r^2$
- A1*: Proceeds using rigorous reasoning to $A=\dfrac{12}{r}+\dfrac{5}{3}\pi r^2$
## Part (b)
| Working | Mark | Guidance |
|---------|------|----------|
| $\left\{A=12r^{-1}+\dfrac{5}{3}\pi r^2\Rightarrow\right\}\ \dfrac{\mathrm{d}A}{\mathrm{d}r}=-12r^{-2}+\dfrac{10}{3}\pi r$ | M1, A1 | 3.4, 1.1b — Differentiates $\frac{\lambda}{r}+\mu r^2$ to give $\alpha r^{-2}+\beta r$ |
| $\left\{\dfrac{\mathrm{d}A}{\mathrm{d}r}=0\Rightarrow\right\}\ -\dfrac{12}{r^2}+\dfrac{10}{3}\pi r=0\Rightarrow-36+10\pi r^3=0\Rightarrow r^{\pm3}=...\left\{=\dfrac{18}{5\pi}\right\}$ | M1 | 2.1 — Sets $\dfrac{\mathrm{d}A}{\mathrm{d}r}=0$ and rearranges to $r^{\pm3}=k$, $k\neq0$ |
| $r=1.046447736...\Rightarrow r=1.05\ \text{(m)}$ (3 sf) or awrt 1.05 (m) | A1 | 1.1b |
**Notes for (b):**
- Note: Give final A1 for correct exact values of $r$, e.g. $r=\left(\dfrac{18}{5\pi}\right)^{\frac{1}{3}}$ or $r=\left(\dfrac{36}{10\pi}\right)^{\frac{1}{3}}$ or $r=\left(\dfrac{3.6}{\pi}\right)^{\frac{1}{3}}$
- Note: Give M0A0M0A0 where $r=1.05$ (m) (3 sf) or awrt 1.05 (m) is found from no working
## Part (c)
| Working | Mark | Guidance |
|---------|------|----------|
| $A_{\min}=\dfrac{12}{(1.046...)}+\dfrac{5}{3}\pi(1.046...)^2$ | M1 | 3.4 |
| $\{A_{\min}=17.20...\Rightarrow\}\ A=17\ (\text{m}^2)$ or $A=$ awrt $17\ (\text{m}^2)$ | A1ft | 1.1b |
## Question 13 (continued):
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Substitutes $r = 1.046...$ (found from solving $\frac{dA}{dr} = 0$ in part (b)) into $A = \frac{12}{r} + \frac{5}{3}\pi r^2$ | M1 | Give M0 for substituting $r$ found from $\frac{d^2A}{dr^2} = 0$ or using $\frac{d^2A}{dr^2}$ into the model |
| $\{A =\}\ 17$ or $\{A =\}$ awrt 17 (ignoring units) | A1ft | Valid only for $0.6 \leq r \leq 1.3$ where $r$ found from $\frac{dA}{dr} = 0$; Give M1A1 for $A = 17\ \text{m}^2$ or $A =$ awrt $17\ \text{m}^2$ from no working |
---13.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{fa4afaf4-fe5d-4f3a-b3de-9600d5502a49-40_501_401_242_831}
\captionsetup{labelformat=empty}
\caption{Figure 9}
\end{center}
\end{figure}
[A sphere of radius $r$ has volume $\frac { 4 } { 3 } \pi r ^ { 3 }$ and surface area $4 \pi r ^ { 2 }$ ]\\
A manufacturer produces a storage tank.\\
The tank is modelled in the shape of a hollow circular cylinder closed at one end with a hemispherical shell at the other end as shown in Figure 9.
The walls of the tank are assumed to have negligible thickness.\\
The cylinder has radius $r$ metres and height $h$ metres and the hemisphere has radius $r$ metres.\\
The volume of the tank is $6 \mathrm {~m} ^ { 3 }$.
\begin{enumerate}[label=(\alph*)]
\item Show that, according to the model, the surface area of the tank, in $\mathrm { m } ^ { 2 }$, is given by
$$\frac { 12 } { r } + \frac { 5 } { 3 } \pi r ^ { 2 }$$
The manufacturer needs to minimise the surface area of the tank.
\item Use calculus to find the radius of the tank for which the surface area is a minimum.\\
(4)
\item Calculate the minimum surface area of the tank, giving your answer to the nearest integer.
\end{enumerate}
\hfill \mbox{\textit{Edexcel Paper 2 2019 Q13 [10]}}