| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2012 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Applied differentiation |
| Type | Optimise 3D shape dimensions |
| Difficulty | Moderate -0.3 This is a standard C2 optimization problem with scaffolded steps. Part (a) is simple rearrangement of V=πr²h, part (b) substitutes into the surface area formula (shown, not derived), parts (c-e) apply routine calculus (differentiate, set to zero, solve, verify). All techniques are textbook exercises 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 |
| \(h = \frac{60}{\pi x^2}\) or equivalent exact expression | B1 | Not decimal; \(\frac{60}{\pi r^2}\) is B0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((A =) 2\pi x^2 + 2\pi xh\) or \((A =) 2\pi r^2 + 2\pi rh\) or \((A =) 2\pi r^2 + \pi dh\) | B1 | Accept any equivalent correct form, may be on separate lines |
| Either \((A) = 2\pi x^2 + 2\pi x\left(\frac{60}{\pi x^2}\right)\) or as \(\pi xh = \frac{60}{x}\) then \((A =) 2\pi x^2 + 2\left(\frac{60}{x}\right)\) | M1 | Substitute expression for \(h\) into area formula of the form \(kx^2 + cxh\) |
| \(A = 2\pi x^2 + \left(\frac{120}{x}\right)\) | A1 cso | At least one power of \(x\) decreased by 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dA}{dx} = 4\pi x - \frac{120}{x^2}\) or \(= 4\pi x - 120x^{-2}\) | M1 A1 | |
| \(4\pi x - \frac{120}{x^2} = 0\) implies \(x^3 =\) ... (Use of \(> 0\) or \(< 0\) is M0 then M0A0) | M1 | Setting \(\frac{dA}{dx} = 0\) and finding value for \(x^3\) |
| \(x = \sqrt[3]{\frac{120}{4\pi}}\) or answers which round to 2.12 (\(-2.12\) is A0) | dM1 A1 | Using cube root to find \(x\). Any equivalent correct answer (3sf or more) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(A = 2\pi(2.12)^2 + \frac{120}{2.12} = 85\) | M1, A1 | Only ft \(x = 2\) or \(x = 2.1\) — both give 85. A1 for 85 only |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Either \(\frac{d^2A}{dx^2} = 4\pi + \frac{240}{x^3}\) and sign considered | M1 | Or (method 2) considers gradient to left and right of 2.12; or (method 3) considers value of \(A\) either side |
| Which is \(> 0\) and therefore minimum | A1 | Clear statements and conclusion. \(A''(x)\) must be correct. Must not see 85 substituted |
## Question 8:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $h = \frac{60}{\pi x^2}$ or equivalent exact expression | B1 | Not decimal; $\frac{60}{\pi r^2}$ is B0 |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(A =) 2\pi x^2 + 2\pi xh$ or $(A =) 2\pi r^2 + 2\pi rh$ or $(A =) 2\pi r^2 + \pi dh$ | B1 | Accept any equivalent correct form, may be on separate lines |
| Either $(A) = 2\pi x^2 + 2\pi x\left(\frac{60}{\pi x^2}\right)$ or as $\pi xh = \frac{60}{x}$ then $(A =) 2\pi x^2 + 2\left(\frac{60}{x}\right)$ | M1 | Substitute expression for $h$ into area formula of the form $kx^2 + cxh$ |
| $A = 2\pi x^2 + \left(\frac{120}{x}\right)$ | A1 cso | At least one power of $x$ decreased by 1 |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dA}{dx} = 4\pi x - \frac{120}{x^2}$ or $= 4\pi x - 120x^{-2}$ | M1 A1 | |
| $4\pi x - \frac{120}{x^2} = 0$ implies $x^3 =$ ... (Use of $> 0$ or $< 0$ is M0 then M0A0) | M1 | Setting $\frac{dA}{dx} = 0$ and finding value for $x^3$ |
| $x = \sqrt[3]{\frac{120}{4\pi}}$ or answers which round to 2.12 ($-2.12$ is A0) | dM1 A1 | Using cube root to find $x$. Any equivalent correct answer (3sf or more) |
### Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $A = 2\pi(2.12)^2 + \frac{120}{2.12} = 85$ | M1, A1 | Only ft $x = 2$ or $x = 2.1$ — both give 85. A1 for **85 only** |
### Part (e):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Either $\frac{d^2A}{dx^2} = 4\pi + \frac{240}{x^3}$ and sign considered | M1 | Or (method 2) considers gradient to left and right of 2.12; or (method 3) considers value of $A$ either side |
| Which is $> 0$ and therefore minimum | A1 | Clear statements and conclusion. $A''(x)$ must be correct. Must **not** see 85 substituted |
---8.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f1ef99f0-4ad4-49d8-bee7-d5bb9cc84660-11_305_446_223_749}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
A manufacturer produces pain relieving tablets. Each tablet is in the shape of a solid circular cylinder with base radius $x \mathrm {~mm}$ and height $h \mathrm {~mm}$, as shown in Figure 3.
Given that the volume of each tablet has to be $60 \mathrm {~mm} ^ { 3 }$,
\begin{enumerate}[label=(\alph*)]
\item express $h$ in terms of $x$,
\item show that the surface area, $A \mathrm {~mm} ^ { 2 }$, of a tablet is given by $A = 2 \pi x ^ { 2 } + \frac { 120 } { x }$
The manufacturer needs to minimise the surface area $A \mathrm {~mm} ^ { 2 }$, of a tablet.
\item Use calculus to find the value of $x$ for which $A$ is a minimum.
\item Calculate the minimum value of $A$, giving your answer to the nearest integer.
\item Show that this value of $A$ is a minimum.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2012 Q8 [13]}}