| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2005 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Applied differentiation |
| Type | Optimise 2D composite shape |
| Difficulty | Standard +0.3 This is a standard optimization problem requiring perimeter constraint setup, area formula derivation, differentiation, and second derivative test. While multi-part, each step follows routine procedures taught in C2 with no novel insights required. The 'show that' in part (a) provides the target formula, making it easier than deriving it independently. Slightly easier than average due to clear structure and standard techniques. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx1.07p Points of inflection: using second derivative |
| Answer | Marks | Guidance |
|---|---|---|
| Area \(\rightarrow A = 2xy + \frac{1}{2}\pi x^2\) | B1, B1 | |
| \(y = \frac{80-2x-\pi x}{2}\) and sub in to \(A\) | M1 | |
| \(\Rightarrow A = 80x - 2x^2 - \pi x^2 + \frac{1}{2}\pi x^2\) | A1 c.s.o | (4 marks) |
| i.e. \(A = 80x - \left(2+\frac{\pi}{2}\right)x^2 *\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dA}{dx} = 0 \Rightarrow 40 = \left(2+\frac{\pi}{2}\right)x\) | M1, A1 | so \(x = \frac{40}{2+\frac{\pi}{2}}\) or \(\frac{80}{4+\pi}\) or Awrt \(11.2\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(< 0\) \(\therefore A\) is Max | M1, A1 | (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(= 448\) (m²) | M1, A1 cao | (2 marks) |
**(a)** Perimeter $\Rightarrow 2x + 2y + \pi x = 80$
Area $\rightarrow A = 2xy + \frac{1}{2}\pi x^2$ | B1, B1 |
$y = \frac{80-2x-\pi x}{2}$ and sub in to $A$ | M1 |
$\Rightarrow A = 80x - 2x^2 - \pi x^2 + \frac{1}{2}\pi x^2$ | A1 c.s.o | (4 marks) |
i.e. $A = 80x - \left(2+\frac{\pi}{2}\right)x^2 *$ |
**(b)** $\frac{dA}{dx} = 80 - 2\left(2+\frac{\pi}{2}\right)x$
$\frac{dA}{dx} = 0 \Rightarrow 40 = \left(2+\frac{\pi}{2}\right)x$ | M1, A1 | so $x = \frac{40}{2+\frac{\pi}{2}}$ or $\frac{80}{4+\pi}$ or Awrt $11.2$ | M1, A1 | (4 marks)
**(c)** $\frac{d^2A}{dx^2} = -4-\pi$
$< 0$ $\therefore A$ is Max | M1, A1 | (2 marks)
**(d)** Max Area $= 80(b) - \left(2+\frac{\pi}{2}\right)(b)^2$
$= 448$ (m²) | M1, A1 cao | (2 marks)
**(Subtotal: 12 marks)**
Guidance: (b) 2nd M1 for putting $\frac{dA}{dx} = 0$ and attempting $x = \cdots$.
(c) M1 for attempting $\frac{d^2A}{dx^2}$ (or equivalent method). A1 for a correct second derivative, $< 0$ and comment.9.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\includegraphics[alt={},max width=\textwidth]{13bca882-27da-40f2-99d8-4fdeb6629c4e-16_821_958_301_516}
\end{center}
\end{figure}
Figure 3 shows the plan of a stage in the shape of a rectangle joined to a semicircle. The length of the rectangular part is $2 x$ metres and the width is $y$ metres. The diameter of the semicircular part is $2 x$ metres. The perimeter of the stage is 80 m .
\begin{enumerate}[label=(\alph*)]
\item Show that the area, $A \mathrm {~m} ^ { 2 }$, of the stage is given by
$$A = 80 x - \left( 2 + \frac { \pi } { 2 } \right) x ^ { 2 } .$$
\item Use calculus to find the value of $x$ at which $A$ has a stationary value.
\item Prove that the value of $x$ you found in part (b) gives the maximum value of $A$.
\item Calculate, to the nearest $\mathrm { m } ^ { 2 }$, the maximum area of the stage.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2005 Q9 [12]}}