| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2014 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Stationary points and optimisation |
| Type | Show formula then optimise: composite/irregular shape |
| Difficulty | Standard +0.3 This is a standard C2 optimization problem with guided steps. Part (a) requires setting up area equations for composite shapes (triangle, rectangle, semicircle), part (b) substitutes into a perimeter formula, and part (c) uses routine differentiation to find a minimum. The algebraic manipulation is straightforward, and part (d) asks for the standard second derivative test. While multi-step, each component is a textbook technique 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 derivatives1.07p Points of inflection: using second derivative |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\{A =\} xy + \frac{\pi}{2}\left(\frac{x}{2}\right)^2 + \frac{1}{2}x^2\sin 60°\) | M1 | An attempt to find 3 areas of the form: \(xy\), \(p\pi x^2\) and \(qx^2\) |
| \(50 = xy + \frac{\pi x^2}{8} + \frac{\sqrt{3}x^2}{4} \Rightarrow y = \frac{50}{x} - \frac{\pi x}{8} - \frac{\sqrt{3}x}{4} \Rightarrow y = \frac{50}{x} - \frac{x}{8}(\pi + 2\sqrt{3})\)* | A1* | Correct expression for \(A\) (terms must be added). Correct proof with no errors seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\{P =\} \frac{\pi x}{2} + 2x + 2y\) | B1 | Correct expression for \(P\) in terms of \(x\) and \(y\) |
| \(P = \frac{\pi x}{2} + 2x + 2\left(\frac{50}{x} - \frac{x}{8}(\pi + 2\sqrt{3})\right)\) | M1 | Substitutes the given expression for \(y\) into an expression for \(P\) where \(P\) is at least of the form \(\alpha x + \beta y\) |
| \(P = \frac{\pi x}{2} + 2x + \frac{100}{x} - \frac{\pi x}{4} - \frac{\sqrt{3}}{2}x \Rightarrow P = \frac{100}{x} + \frac{\pi x}{4} + 2x - \frac{\sqrt{3}}{2}x\) | ||
| \(\Rightarrow P = \frac{100}{x} + \frac{x}{4}(\pi + 8 - 2\sqrt{3})\)* | A1* | Correct proof with no errors seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dP}{dx} = -100x^{-2} + \frac{\pi + 8 - 2\sqrt{3}}{4}\) | M1A1 | M1: Either \(\mu x \to \mu\) or \(\frac{100}{x} \to \frac{\pm\lambda}{x^2}\). A1: Correct differentiation (need not be simplified). Allow \(-100x^{-2} + (\text{awrt } 1.92)\) |
| \(-100x^{-2} + \frac{\pi + 8 - 2\sqrt{3}}{4} = 0 \Rightarrow x = \ldots\) | M1 | Their \(P' = 0\) and attempt to solve as far as \(x = \ldots\) (ignore poor manipulation) |
| \(\Rightarrow x = \sqrt{\frac{400}{\pi + 8 - 2\sqrt{3}}} = 7.2180574\ldots\) | A1 | \(\sqrt{\frac{400}{\pi + 8 - 2\sqrt{3}}}\) or awrt 7.2 and no other values |
| \(\{x = 7.218\ldots\} \Rightarrow P = 27.708\ldots\) (m) | A1 | awrt 27.7 |
| \(\frac{d^2P}{dx^2} = \frac{200}{x^3} > 0 \Rightarrow\) Minimum | M1A1ft | M1: Finds \(P''\) (\(x^n \to x^{n-1}\), allow for constant \(\to 0\)) and considers sign. A1ft: \(\frac{200}{x^3}\) (need not be simplified) and \(> 0\) and conclusion. Only follow through on correct \(P''\) and single positive \(x\) |
## Question 9:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\{A =\} xy + \frac{\pi}{2}\left(\frac{x}{2}\right)^2 + \frac{1}{2}x^2\sin 60°$ | M1 | An attempt to find 3 areas of the form: $xy$, $p\pi x^2$ and $qx^2$ |
| $50 = xy + \frac{\pi x^2}{8} + \frac{\sqrt{3}x^2}{4} \Rightarrow y = \frac{50}{x} - \frac{\pi x}{8} - \frac{\sqrt{3}x}{4} \Rightarrow y = \frac{50}{x} - \frac{x}{8}(\pi + 2\sqrt{3})$* | A1* | Correct expression for $A$ (terms must be added). Correct proof with no errors seen |
**[3 marks]**
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\{P =\} \frac{\pi x}{2} + 2x + 2y$ | B1 | Correct expression for $P$ in terms of $x$ and $y$ |
| $P = \frac{\pi x}{2} + 2x + 2\left(\frac{50}{x} - \frac{x}{8}(\pi + 2\sqrt{3})\right)$ | M1 | Substitutes the given expression for $y$ into an expression for $P$ where $P$ is at least of the form $\alpha x + \beta y$ |
| $P = \frac{\pi x}{2} + 2x + \frac{100}{x} - \frac{\pi x}{4} - \frac{\sqrt{3}}{2}x \Rightarrow P = \frac{100}{x} + \frac{\pi x}{4} + 2x - \frac{\sqrt{3}}{2}x$ | | |
| $\Rightarrow P = \frac{100}{x} + \frac{x}{4}(\pi + 8 - 2\sqrt{3})$* | A1* | Correct proof with no errors seen |
**[3 marks]**
### Parts (c) and (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dP}{dx} = -100x^{-2} + \frac{\pi + 8 - 2\sqrt{3}}{4}$ | M1A1 | M1: Either $\mu x \to \mu$ or $\frac{100}{x} \to \frac{\pm\lambda}{x^2}$. A1: Correct differentiation (need not be simplified). Allow $-100x^{-2} + (\text{awrt } 1.92)$ |
| $-100x^{-2} + \frac{\pi + 8 - 2\sqrt{3}}{4} = 0 \Rightarrow x = \ldots$ | M1 | Their $P' = 0$ and attempt to solve as far as $x = \ldots$ (ignore poor manipulation) |
| $\Rightarrow x = \sqrt{\frac{400}{\pi + 8 - 2\sqrt{3}}} = 7.2180574\ldots$ | A1 | $\sqrt{\frac{400}{\pi + 8 - 2\sqrt{3}}}$ or awrt 7.2 **and no other values** |
| $\{x = 7.218\ldots\} \Rightarrow P = 27.708\ldots$ (m) | A1 | awrt 27.7 |
| $\frac{d^2P}{dx^2} = \frac{200}{x^3} > 0 \Rightarrow$ Minimum | M1A1ft | M1: Finds $P''$ ($x^n \to x^{n-1}$, allow for constant $\to 0$) and considers sign. A1ft: $\frac{200}{x^3}$ (need not be simplified) and $> 0$ and conclusion. Only follow through on correct $P''$ and single positive $x$ |
**[5+2 marks]**
---9.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f07cc9ed-a820-46c8-a3a3-3c780cf20fa7-14_899_686_212_639}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
Figure 4 shows the plan of a pool.
The shape of the pool $A B C D E F A$ consists of a rectangle $B C E F$ joined to an equilateral triangle $B F A$ and a semi-circle $C D E$, as shown in Figure 4.
Given that $A B = x$ metres, $E F = y$ metres, and the area of the pool is $50 \mathrm {~m} ^ { 2 }$,
\begin{enumerate}[label=(\alph*)]
\item show that
$$y = \frac { 50 } { x } - \frac { x } { 8 } ( \pi + 2 \sqrt { } 3 )$$
\item Hence show that the perimeter, $P$ metres, of the pool is given by
$$P = \frac { 100 } { x } + \frac { x } { 4 } ( \pi + 8 - 2 \sqrt { } 3 )$$
\item Use calculus to find the minimum value of $P$, giving your answer to 3 significant figures.
\item Justify, by further differentiation, that the value of $P$ that you have found is a minimum.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2014 Q9 [13]}}