| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2015 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Applied differentiation |
| Type | Optimise 3D shape dimensions |
| Difficulty | Standard +0.3 This is a standard optimization problem with guided steps. Parts (a) and (b) involve straightforward algebraic manipulation using given formulas. Part (c) requires routine differentiation and solving dA/dR = 0. Part (d) uses the second derivative test. Part (e) is simple substitution. While multi-part, each step follows a predictable template with no novel insight required, making it slightly easier than average. |
| Spec | 1.02z Models in context: use functions in modelling |
| Answer | Marks | Guidance |
|---|---|---|
| \(\pi R^2 H + \frac{2}{3}\pi R^3 = 800\pi\) so \(H = \frac{800}{R^2} - \frac{2}{3}R\) | M1, A1* | Sets up volume equation; show that — intermediate step required showing division of \(\pi r^2 / \pi R^2\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(A = \pi R^2 + 2\pi RH + 2\pi R^2\) | B1 | Correct expression with three separate correct elements; allow \(A = \pi R^2 + 2\pi RH + \frac{4\pi R^2}{2}\) |
| \(A = 3\pi R^2 + 2\pi R\left(\frac{800}{R^2} - \frac{2}{3}R\right)\) so \(A = \frac{5\pi R^2}{3} + \frac{1600\pi}{R}\) | M1, A1* | Score for substituting \(H\); show that — all aspects must be correct |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dA}{dR} = \frac{10}{3}\pi R - \frac{1600\pi}{R^2}\) | M1, A1 | Evidence of differentiation: \(R^2 \to R\) or \(R^{-1} \to R^{-2}\); both terms correct |
| Set \(\frac{dA}{dR} = 0\) and obtain \(R^3 = 480\) | dM1, A1 | Dependent on previous M; correct intermediate answer |
| \(R = 7.83\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{d^2A}{dR^2} = \frac{10\pi}{3} + 3200\pi R^{-3} > 0\) so minimum | M1A1 | Correct second derivative; or substitute positive \(R\) into second derivative; or check gradient either side of \(R\); correct conclusion required |
| Answer | Marks |
|---|---|
| \(H = \) awrt \(7.83\) | B1 |
# Question 16:
## Part (a):
| $\pi R^2 H + \frac{2}{3}\pi R^3 = 800\pi$ so $H = \frac{800}{R^2} - \frac{2}{3}R$ | M1, A1* | Sets up volume equation; show that — intermediate step required showing division of $\pi r^2 / \pi R^2$ |
## Part (b):
| $A = \pi R^2 + 2\pi RH + 2\pi R^2$ | B1 | Correct expression with three separate correct elements; allow $A = \pi R^2 + 2\pi RH + \frac{4\pi R^2}{2}$ |
|---|---|---|
| $A = 3\pi R^2 + 2\pi R\left(\frac{800}{R^2} - \frac{2}{3}R\right)$ so $A = \frac{5\pi R^2}{3} + \frac{1600\pi}{R}$ | M1, A1* | Score for substituting $H$; show that — all aspects must be correct |
## Part (c):
| $\frac{dA}{dR} = \frac{10}{3}\pi R - \frac{1600\pi}{R^2}$ | M1, A1 | Evidence of differentiation: $R^2 \to R$ or $R^{-1} \to R^{-2}$; both terms correct |
|---|---|---|
| Set $\frac{dA}{dR} = 0$ and obtain $R^3 = 480$ | dM1, A1 | Dependent on previous M; correct intermediate answer |
| $R = 7.83$ | **A1** | |
## Part (d):
| $\frac{d^2A}{dR^2} = \frac{10\pi}{3} + 3200\pi R^{-3} > 0$ so minimum | M1A1 | Correct second derivative; or substitute positive $R$ into second derivative; or check gradient either side of $R$; correct conclusion required |
## Part (e):
| $H = $ awrt $7.83$ | B1 | |\begin{enumerate}
\item \hspace{0pt} [In this question you may assume the formula for the area of a circle and the following formulae:\\
a sphere of radius $r$ has volume $V = \frac { 4 } { 3 } \pi r ^ { 3 }$ and surface area $S = 4 \pi r ^ { 2 }$\\
a cylinder of radius $r$ and height $h$ has volume $V = \pi r ^ { 2 } h$ and curved surface area $S = 2 \pi r h ]$
\end{enumerate}
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{ea81408b-e292-4529-b1e2-e3246503a3ac-25_414_478_566_726}
\captionsetup{labelformat=empty}
\caption{Figure 5}
\end{center}
\end{figure}
Figure 5 shows the model for a building. The model is made up of three parts. The roof is modelled by the curved surface of a hemisphere of radius $R \mathrm {~cm}$. The walls are modelled by the curved surface of a circular cylinder of radius $R \mathrm {~cm}$ and height $H \mathrm {~cm}$. The floor is modelled by a circular disc of radius $R \mathrm {~cm}$. The model is made of material of negligible thickness, and the walls are perpendicular to the base.
It is given that the volume of the model is $800 \pi \mathrm {~cm} ^ { 3 }$ and that $0 < R < 10.6$\\
(a) Show that
$$H = \frac { 800 } { R ^ { 2 } } - \frac { 2 } { 3 } R$$
(b) Show that the surface area, $A \mathrm {~cm} ^ { 2 }$, of the model is given by
$$A = \frac { 5 \pi R ^ { 2 } } { 3 } + \frac { 1600 \pi } { R }$$
(c) Use calculus to find the value of $R$, to 3 significant figures, for which $A$ is a minimum.\\
(d) Prove that this value of $R$ gives a minimum value for $A$.\\
(e) Find, to 3 significant figures, the value of $H$ which corresponds to this value for $R$.
\hfill \mbox{\textit{Edexcel C12 2015 Q16 [13]}}