| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2016 |
| Session | October |
| Marks | 13 |
| 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 surface area constraint manipulation, substitution to get V(r), differentiation using quotient/product rule, and second derivative test. While multi-step, each component is routine Core 1/2 material with no novel insight required—slightly easier than average A-level questions. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.07p Points of inflection: using second derivative1.07t Construct differential equations: in context |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(200 = \pi r^2 + \pi rh + 2rh\) | M1 A1 | M1: sets total surface area equal to \(200\) with at least two correct terms; note \(200 = 2\pi r^2 + \pi rh\) does not mean two terms are correct |
| \((h =)\frac{200-\pi r^2}{\pi r + 2r}\) or \((rh =)\frac{200-\pi r^2}{\pi + 2}\) | dM1 | Makes \(h\) or \(rh\) the subject; must have had two terms in \(h\); dependent on previous M1 |
| \(V = \frac{1}{2}\pi r^2 h\) | M1 | May be implied by sight of \(V = \frac{1}{2}\pi r^2 \times \text{their } h\) |
| \(\Rightarrow V = \frac{\pi r^2(200-\pi r^2)}{2(2r+\pi r)} = \frac{\pi r(200-\pi r^2)}{4+2\pi}\) | A1* | cso — substitutes for \(r\) or \(rh\) correctly and proceeds correctly |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dV}{dr} = \frac{200\pi - 3\pi^2 r^2}{4+2\pi}\) | M1 A1 | M1: attempts to differentiate \(V\) or numerator of \(V\); accept \(\frac{dV}{dr} = A \pm Br^2\); also accept awrt \(\frac{dV}{dr} = 61.1 - 2.9r^2\); may see \((4+2\pi)\frac{dV}{dr} = A \pm Br^2\) |
| \(\frac{200\pi - 3\pi^2 r^2}{4+2\pi} = 0\) or \(200\pi - 3\pi^2 r^2 = 0\) leading to \(r^2 =\) | dM1 | Setting \(\frac{dV}{dr}=0\) and finding value for \(r^2\); may be implied by answer |
| \(r = \sqrt{\frac{200}{3\pi}}\) or answers which round to \(4.6\) | dM1 A1 | Using square root to find \(r\); dependent on all previous M marks; correct answer implies previous two M marks; accept awrt \(4.6\) |
| \(V = 188\) | B1 | Exact answer only; do not accept e.g. \(187.8\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{d^2V}{dr^2} = \frac{-6\pi^2 r}{4+2\pi}\), and sign considered | M1 | Score for second derivative \(= \pm Cr\) and considers the sign; accept \(\frac{d^2V}{dr^2} = \text{awrt} -5.8r\); can be implied by \(\frac{\pi r(200-\pi r^2)}{4+2\pi} \to A \pm Br^2 \to \pm Cr\) with consideration of sign |
| \(\left.\frac{d^2V}{dr^2}\right\ | _{r=\ldots} = -27 < 0\) and therefore maximum | A1 |
# Question 15(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $200 = \pi r^2 + \pi rh + 2rh$ | M1 A1 | M1: sets total surface area equal to $200$ with at least two correct terms; note $200 = 2\pi r^2 + \pi rh$ does not mean two terms are correct |
| $(h =)\frac{200-\pi r^2}{\pi r + 2r}$ or $(rh =)\frac{200-\pi r^2}{\pi + 2}$ | dM1 | Makes $h$ or $rh$ the subject; must have had two terms in $h$; dependent on previous M1 |
| $V = \frac{1}{2}\pi r^2 h$ | M1 | May be implied by sight of $V = \frac{1}{2}\pi r^2 \times \text{their } h$ |
| $\Rightarrow V = \frac{\pi r^2(200-\pi r^2)}{2(2r+\pi r)} = \frac{\pi r(200-\pi r^2)}{4+2\pi}$ | A1* | cso — substitutes for $r$ or $rh$ correctly and proceeds correctly |
---
# Question 15(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dV}{dr} = \frac{200\pi - 3\pi^2 r^2}{4+2\pi}$ | M1 A1 | M1: attempts to differentiate $V$ or numerator of $V$; accept $\frac{dV}{dr} = A \pm Br^2$; also accept awrt $\frac{dV}{dr} = 61.1 - 2.9r^2$; may see $(4+2\pi)\frac{dV}{dr} = A \pm Br^2$ |
| $\frac{200\pi - 3\pi^2 r^2}{4+2\pi} = 0$ or $200\pi - 3\pi^2 r^2 = 0$ leading to $r^2 =$ | dM1 | Setting $\frac{dV}{dr}=0$ and finding value for $r^2$; may be implied by answer |
| $r = \sqrt{\frac{200}{3\pi}}$ or answers which round to $4.6$ | dM1 A1 | Using square root to find $r$; dependent on all previous M marks; correct answer implies previous two M marks; accept awrt $4.6$ |
| $V = 188$ | B1 | Exact answer only; do not accept e.g. $187.8$ |
---
# Question 15(c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{d^2V}{dr^2} = \frac{-6\pi^2 r}{4+2\pi}$, and sign considered | M1 | Score for second derivative $= \pm Cr$ and considers the sign; accept $\frac{d^2V}{dr^2} = \text{awrt} -5.8r$; can be implied by $\frac{\pi r(200-\pi r^2)}{4+2\pi} \to A \pm Br^2 \to \pm Cr$ with consideration of sign |
| $\left.\frac{d^2V}{dr^2}\right\|_{r=\ldots} = -27 < 0$ and therefore maximum | A1 | Clear statements and conclusion required: (1) $\frac{d^2V}{dr^2}$ must be correct (not just numerator); (2) statement that when their $r$ is substituted, $\frac{d^2V}{dr^2}$ is negative or $< 0$; (3) minimal conclusion such as "hence maximum"; e.g. accept $\frac{d^2V}{dr^2} = -5.76r$, when $r=4.5 \Rightarrow \frac{d^2V}{dr^2} < 0$, hence max |15.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{53865e15-3838-4551-b507-fe49549b87db-44_647_917_260_484}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
Figure 4 shows a solid wooden block. The block is a right prism with length $h \mathrm {~cm}$. The cross-section of the block is a semi-circle with radius $r \mathrm {~cm}$.
The total surface area of the block, including the curved surface, the two semi-circular ends and the rectangular base, is $200 \mathrm {~cm} ^ { 2 }$
\begin{enumerate}[label=(\alph*)]
\item Show that the volume $V \mathrm {~cm} ^ { 3 }$ of the block is given by
$$V = \frac { \pi r \left( 200 - \pi r ^ { 2 } \right) } { 4 + 2 \pi }$$
\item Use calculus to find the maximum value of $V$. Give your answer to the nearest $\mathrm { cm } ^ { 3 }$.
\item Justify, by further differentiation, that the value of $V$ that you have found is a maximum.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2016 Q15 [13]}}