| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Session | Specimen |
| Marks | 11 |
| 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 optimization problem requiring area constraint manipulation, perimeter formulation, and basic differentiation to find a minimum. Parts (a) and (b) are guided 'show that' steps requiring straightforward algebra, while part (c) involves routine calculus (dP/dx = 0). The problem is slightly easier than average as it's highly scaffolded with no geometric insight required beyond reading the diagram. |
| Spec | 1.02z Models in context: use functions in modelling1.07t Construct differential equations: in context |
13.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{1528bec3-7a7a-42c5-bac2-756ff3493818-28_374_410_278_776}
\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 $P$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 Q13 [11]}}