| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2009 |
| Session | January |
| Marks | 12 |
| 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 C2 optimization problem with scaffolding. Part (a) is given (show that), requiring only algebraic manipulation of surface area formula. Parts (b) and (c) follow routine calculus procedures: differentiate, set to zero, solve, and verify maximum using second derivative test. All steps are textbook-standard with no novel insight 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 |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(2\pi rh + 2\pi r^2 = 800\) | B1 | For any correct form of this equation (may be unsimplified, may be implied by 1st M1) |
| \(h = \frac{400 - \pi r^2}{\pi r}\) | M1 | Making \(h\) the subject of their three or four term formula |
| \(V = \pi r^2\left(\frac{400 - \pi r^2}{\pi r}\right)\) | M1 | Substituting expression for \(h\) into \(\pi r^2 h\) (independent mark). Must now be expression in \(r\) only |
| \(V = 400r - \pi r^3\) (*) | A1 | cso |
| (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dV}{dr} = 400 - 3\pi r^2\) | M1 A1 | M1: At least one power of \(r\) decreased by 1. A1: cao |
| \(400 - 3\pi r^2 = 0 \implies r^2 = \ldots \implies r = \sqrt{\frac{400}{3\pi}}\) \((= 6.5\) (2 s.f.)) | M1 A1 | M1: Setting \(\frac{dV}{dr} = 0\) and finding a value for correct power of \(r\). A1: This mark may be credited if value of \(V\) is correct. Otherwise round to 6.5 (allow \(\pm 6.5\)) or be exact answer |
| \(V = 400r - \pi r^3 = 1737 = \frac{800}{3}\sqrt{\frac{400}{3\pi}}\ (\text{cm}^3)\) | M1 A1 | M1: Substitute a positive value of \(r\) to give \(V\). A1: 1737 or 1737.25… or exact answer |
| (accept awrt 1737 or exact answer) | ||
| (6) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{d^2V}{dr^2} = -6\pi r\), Negative, \(\therefore\) maximum | M1 A1 | M1: needs complete method e.g. attempts differentiation (power reduced) of their first derivative and considers its sign. A1 (first method): should be \(-6\pi r\) (do not need to substitute \(r\), can condone wrong \(r\) if found in (b)). Need to conclude maximum or indicate by a tick that it is maximum. Throughout allow confused notation such as \(dy/dx\) for \(dV/dr\) |
| (2) |
# Question 10:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $2\pi rh + 2\pi r^2 = 800$ | B1 | For any correct form of this equation (may be unsimplified, may be implied by 1st M1) |
| $h = \frac{400 - \pi r^2}{\pi r}$ | M1 | Making $h$ the subject of their three or four term formula |
| $V = \pi r^2\left(\frac{400 - \pi r^2}{\pi r}\right)$ | M1 | Substituting expression for $h$ into $\pi r^2 h$ (independent mark). Must now be expression in $r$ only |
| $V = 400r - \pi r^3$ (*) | A1 | cso |
| | **(4)** | |
**Alternative for (a):** $A = 2\pi r^2 + 2\pi rh$, $\frac{A}{2} \times r = \pi r^3 + \pi r^2 h$ is **M1**. Equate to $400r$ **B1**. Then $V = 400r - \pi r^3$ is **M1 A1**
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dV}{dr} = 400 - 3\pi r^2$ | M1 A1 | M1: At least one power of $r$ decreased by 1. A1: cao |
| $400 - 3\pi r^2 = 0 \implies r^2 = \ldots \implies r = \sqrt{\frac{400}{3\pi}}$ $(= 6.5$ (2 s.f.)) | M1 A1 | M1: Setting $\frac{dV}{dr} = 0$ and finding a value for correct power of $r$. A1: This mark may be credited if value of $V$ is correct. Otherwise round to 6.5 (allow $\pm 6.5$) or be exact answer |
| $V = 400r - \pi r^3 = 1737 = \frac{800}{3}\sqrt{\frac{400}{3\pi}}\ (\text{cm}^3)$ | M1 A1 | M1: Substitute a positive value of $r$ to give $V$. A1: 1737 or 1737.25… or exact answer |
| (accept awrt 1737 or exact answer) | | |
| | **(6)** | |
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{d^2V}{dr^2} = -6\pi r$, Negative, $\therefore$ maximum | M1 A1 | M1: needs complete method e.g. attempts differentiation (power reduced) of their first derivative and considers its sign. A1 (first method): should be $-6\pi r$ (do not need to substitute $r$, can condone wrong $r$ if found in (b)). Need to conclude maximum or indicate by a tick that it is maximum. Throughout allow confused notation such as $dy/dx$ for $dV/dr$ |
| | **(2)** | |
**Other methods for part (c):**
- Either: M: Find value of $\frac{dV}{dr}$ on each side of $r = \sqrt{\frac{400}{3\pi}}$ and consider sign. A: Indicate sign change of positive to negative for $\frac{dV}{dr}$, and conclude max.
- Or: M: Find value of $V$ on each side of $r = \sqrt{\frac{400}{3\pi}}$ and compare with "1737". A: Indicate that both values are less than 1737 or 1737.25, and conclude max.
**Total: [12]**10. A solid right circular cylinder has radius $r \mathrm {~cm}$ and height $h \mathrm {~cm}$.
The total surface area of the cylinder is $800 \mathrm {~cm} ^ { 2 }$.
\begin{enumerate}[label=(\alph*)]
\item Show that the volume, $V \mathrm {~cm} ^ { 3 }$, of the cylinder is given by
$$V = 400 r - \pi r ^ { 3 }$$
Given that $r$ varies,
\item use calculus to find the maximum value of $V$, to the nearest $\mathrm { cm } ^ { 3 }$.
\item Justify that the value of $V$ you have found is a maximum.\\
\includegraphics[max width=\textwidth, alt={}, center]{12e54724-64a3-4dc0-b7d5-6ef6cc04124c-16_103_63_2477_1873}
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2009 Q10 [12]}}