| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2012 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Stationary points and optimisation |
| Type | Optimise perimeter or area of 2D region |
| Difficulty | Standard +0.3 This is a standard C2 optimization problem with guided steps. Part (a) requires setting up an area equation (quarter circle plus two rectangles), part (b) involves algebraic manipulation to express perimeter in terms of one variable, and part (c) uses routine differentiation to find a minimum. The question provides significant scaffolding and uses familiar techniques without requiring novel insight or complex multi-step reasoning. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.07t Construct differential equations: in context |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(kr^2 + cxy = 4\) or \(kr^2 + c[(x+y)^2 - x^2 - y^2] = 4\) | M1 | \(k\) and \(c\) may be wrong |
| \(\frac{1}{4}\pi x^2 + 2xy = 4\) | A1 | Any correct form with \(x\) for radius |
| \(y = \frac{4 - \frac{1}{4}\pi x^2}{2x} = \frac{16 - \pi x^2}{8x}\) | B1 | Making \(y\) subject; no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P = 2x + cy + k\pi r\) where \(c=2\) or \(4\), \(k = \frac{1}{4}\) or \(\frac{1}{2}\) | M1 | Uses correct perimeter formula form |
| \(P = \frac{\pi x}{2} + 2x + 4\left(\frac{4 - \frac{1}{4}\pi x^2}{2x}\right)\) | A1 | Correct unsimplified with \(y\) substituted |
| \(P = \frac{\pi x}{2} + 2x + \frac{8}{x} - \frac{\pi x}{2}\), so \(P = \frac{8}{x} + 2x\) | A1 | Printed answer; at least one line of simplification shown |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dP}{dx} = -\frac{8}{x^2} + 2\) | M1, A1 | At least one power of \(x\) decreased by 1 |
| \(-\frac{8}{x^2} + 2 = 0 \Rightarrow x^2 = 4\), so \(x = 2\) | M1, A1 | Setting \(\frac{dP}{dx}=0\); ignore \(x=-2\) |
| \(P = 4 + 4 = 8\) (m) | B1 | cao; may be implied by correct \(P\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(y = \frac{4-\pi}{4}\), width \(= 21\) (cm) | M1, A1 | Substitute \(x=2\) into \(y\) from (a); accept 21 or 21 cm or 0.21 m |
## Question 8:
### Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $kr^2 + cxy = 4$ or $kr^2 + c[(x+y)^2 - x^2 - y^2] = 4$ | M1 | $k$ and $c$ may be wrong |
| $\frac{1}{4}\pi x^2 + 2xy = 4$ | A1 | Any correct form with $x$ for radius |
| $y = \frac{4 - \frac{1}{4}\pi x^2}{2x} = \frac{16 - \pi x^2}{8x}$ | B1 | Making $y$ subject; no errors |
### Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P = 2x + cy + k\pi r$ where $c=2$ or $4$, $k = \frac{1}{4}$ or $\frac{1}{2}$ | M1 | Uses correct perimeter formula form |
| $P = \frac{\pi x}{2} + 2x + 4\left(\frac{4 - \frac{1}{4}\pi x^2}{2x}\right)$ | A1 | Correct unsimplified with $y$ substituted |
| $P = \frac{\pi x}{2} + 2x + \frac{8}{x} - \frac{\pi x}{2}$, so $P = \frac{8}{x} + 2x$ | A1 | Printed answer; at least one line of simplification shown |
### Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dP}{dx} = -\frac{8}{x^2} + 2$ | M1, A1 | At least one power of $x$ decreased by 1 |
| $-\frac{8}{x^2} + 2 = 0 \Rightarrow x^2 = 4$, so $x = 2$ | M1, A1 | Setting $\frac{dP}{dx}=0$; ignore $x=-2$ |
| $P = 4 + 4 = 8$ (m) | B1 | cao; may be implied by correct $P$ |
### Part (d)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y = \frac{4-\pi}{4}$, width $= 21$ (cm) | M1, A1 | Substitute $x=2$ into $y$ from (a); accept 21 or 21 cm or 0.21 m |
---8.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{42116a65-60ec-4dff-a05e-bab529939e1e-11_403_440_262_744}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Figure 3 shows a flowerbed. Its shape is a quarter of a circle of radius $x$ metres with two equal rectangles attached to it along its radii. Each rectangle has length equal to $x$ metres and width equal to $y$ metres.
Given that the area of the flowerbed is $4 \mathrm {~m} ^ { 2 }$,
\begin{enumerate}[label=(\alph*)]
\item show that
$$y = \frac { 16 - \pi x ^ { 2 } } { 8 x }$$
\item Hence show that the perimeter $P$ metres of the flowerbed is given by the equation
$$P = \frac { 8 } { x } + 2 x$$
\item Use calculus to find the minimum value of $P$.
\item Find the width of each rectangle when the perimeter is a minimum. Give your answer to the nearest centimetre.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2012 Q8 [13]}}