| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2016 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Stationary points and optimisation |
| Type | Show formula then optimise: cylinder/prism (single variable) |
| Difficulty | Standard +0.3 This is a standard optimization problem requiring volume constraint substitution, differentiation of a rational function, and second derivative test. While it involves multiple steps, each technique is routine for C1/C2 level with no novel insight required—slightly easier than average due to straightforward setup and calculus. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.08a Fundamental theorem of calculus: integration as reverse of differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(60000 = \pi r^2 h \Rightarrow h = \frac{60000}{\pi r^2}\) | M1 | Uses \(60000 = \pi r^2 h \Rightarrow h = ..\) Condone errors on number of zeros but formula must be correct |
| Subs in \(S = \pi r^2 + 2\pi r h \Rightarrow S = \pi r^2 + 2\pi r \times \frac{60000}{\pi r^2}\) | M1 | Score for attempt to substitute \(h = ..\) or \(\pi r h = ..\) from dimensionally correct formula for \(V\) into \(S = k\pi r^2 + 2\pi r h\) where \(k = 1\) or \(2\) |
| \(S = \pi r^2 + \frac{120000}{r}\) | A1* | Completes proof with no errors or omissions; allow from \(S = \pi r^2 + \frac{2V}{r}\) if quoted; \(S =\) must appear somewhere in proof |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\frac{dS}{dr} = 2\pi r - \frac{120000}{r^2}\) | M1, A1 | Differentiate the given function and get one power correct; condone \(\frac{dy}{dx}\) throughout (b) |
| \(\frac{dS}{dr} = 0 \Rightarrow r^3 = \frac{120000}{2\pi} \Rightarrow r = \text{awrt } 27 \text{ (cm)}\) | dM1, A1 | Sets \(\frac{dS}{dr} = 0\), multiplies by \(r^2\) and proceeds with correct method to \(r^3 = ..\); \(r = \text{awrt } 27\) or \(r = \sqrt[3]{\frac{120000}{2\pi}}\) |
| \(S = \pi \times 26.7^2 + \frac{120000}{26.7} = \text{awrt } 6730 \text{ (cm}^2\text{)}\) | dM1, A1 | Substitutes positive \(r\) into \(S = \pi r^2 + \frac{120000}{r}\); awrt \(3sf\) \(6730 \text{ cm}^2\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\frac{d^2S}{dr^2} = 2\pi + \frac{240000}{r^3}\bigg | _{r=26.7} = \text{awrt } 19 > 0 \Rightarrow \text{Minimum}\) | M1, A1 |
## Question 15:
### Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| $60000 = \pi r^2 h \Rightarrow h = \frac{60000}{\pi r^2}$ | M1 | Uses $60000 = \pi r^2 h \Rightarrow h = ..$ Condone errors on number of zeros but formula must be correct |
| Subs in $S = \pi r^2 + 2\pi r h \Rightarrow S = \pi r^2 + 2\pi r \times \frac{60000}{\pi r^2}$ | M1 | Score for attempt to substitute $h = ..$ or $\pi r h = ..$ from dimensionally correct formula for $V$ into $S = k\pi r^2 + 2\pi r h$ where $k = 1$ or $2$ |
| $S = \pi r^2 + \frac{120000}{r}$ | A1* | Completes proof with no errors or omissions; allow from $S = \pi r^2 + \frac{2V}{r}$ if quoted; $S =$ must appear somewhere in proof |
### Part (b):
| Working | Mark | Guidance |
|---------|------|----------|
| $\frac{dS}{dr} = 2\pi r - \frac{120000}{r^2}$ | M1, A1 | Differentiate the given function and get one power correct; condone $\frac{dy}{dx}$ throughout (b) |
| $\frac{dS}{dr} = 0 \Rightarrow r^3 = \frac{120000}{2\pi} \Rightarrow r = \text{awrt } 27 \text{ (cm)}$ | dM1, A1 | Sets $\frac{dS}{dr} = 0$, multiplies by $r^2$ and proceeds with correct method to $r^3 = ..$; $r = \text{awrt } 27$ or $r = \sqrt[3]{\frac{120000}{2\pi}}$ |
| $S = \pi \times 26.7^2 + \frac{120000}{26.7} = \text{awrt } 6730 \text{ (cm}^2\text{)}$ | dM1, A1 | Substitutes positive $r$ into $S = \pi r^2 + \frac{120000}{r}$; awrt $3sf$ $6730 \text{ cm}^2$ |
### Part (c):
| Working | Mark | Guidance |
|---------|------|----------|
| $\frac{d^2S}{dr^2} = 2\pi + \frac{240000}{r^3}\bigg|_{r=26.7} = \text{awrt } 19 > 0 \Rightarrow \text{Minimum}$ | M1, A1 | Subs positive $r$ from (b) into $\frac{d^2S}{dr^2} = A \pm \frac{B}{r^3}$, $A,B \neq 0$ and calculates value; $\frac{d^2S}{dr^2} = 2\pi + \frac{240000}{r^3}$ with correct substitution and statement $\frac{d^2S}{dr^2} > 0 \Rightarrow$ Minimum |
---15.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{aa75f1c1-ee97-4fee-af98-957e6a3fbba1-23_609_493_223_762}
\captionsetup{labelformat=empty}
\caption{Figure 5}
\end{center}
\end{figure}
Figure 5 shows a design for a water barrel.\\
It is in the shape of a right circular cylinder with height $h \mathrm {~cm}$ and radius $r \mathrm {~cm}$.
The barrel has a base but has no lid, is open at the top and is made of material of negligible thickness.
The barrel is designed to hold $60000 \mathrm {~cm} ^ { 3 }$ of water when full.
\begin{enumerate}[label=(\alph*)]
\item Show that the total external surface area, $S \mathrm {~cm} ^ { 2 }$, of the barrel is given by the formula
$$S = \pi r ^ { 2 } + \frac { 120000 } { r }$$
\item Use calculus to find the minimum value of $S$, giving your answer to 3 significant figures.
\item Justify that the value of $S$ you found in part (b) is a minimum.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2016 Q15 [11]}}