| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2024 |
| Session | June |
| 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 AS-level optimization problem with clearly defined steps: deriving a constraint equation (area = 100), forming the perimeter function, finding stationary points, and verifying it's a minimum. The geometry is straightforward (quarter circle + rectangles), and part (d) simply requires checking if x ≥ 1. All techniques are routine for AS calculus with no novel insight required, making it slightly easier than average. |
| Spec | 1.02z Models in context: use functions in modelling1.08f Area between two curves: using integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Surface area \(= 2xy + \frac{\pi x^2}{4}\) | B1 | 1.1b – Correct expression for surface area in terms of \(x\) and \(y\) only |
| \(2xy + \frac{\pi x^2}{4} = 100 \Rightarrow y = \frac{400 - \pi x^2}{8x} = \frac{50}{x} - \frac{\pi x}{8}\) | M1 | 3.4 – Sets expression equal to 100 and rearranges to make \(y\) subject |
| \((P =) 2x + 4y + \frac{2\pi x}{4}\) | B1 | 1.1b – Correct expression for perimeter in terms of \(x\) and \(y\) |
| \((P =) 2x + 4\left(\frac{400 - \pi x^2}{8x}\right) + \frac{2\pi x}{4}\) | M1 | 3.4 – Substitutes their \(y\) into perimeter expression |
| \(P = 2x + \frac{200}{x}\) * | A1* | 2.1 – cso. Condone omission of \(P=\) on final line if seen earlier |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dP}{dx} = 2 - 200x^{-2}\) | M1, A1 | 1.1b – Attempts differentiation achieving form \(A \pm Bx^{-2}\); correct answer |
| \(2 - 200x^{-2} = 0 \Rightarrow x = \ldots\) | dM1 | 3.1b – Sets derivative of form \(A - Bx^{-2}\) equal to 0, rearranges. Dependent on M1 |
| \(x = 10\) | A1 | 1.1b – cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{d^2P}{dx^2} = \text{"400"}x^{-3} \Rightarrow \text{"400"} \times 10^{-3} > 0\) | M1 | 1.1b – Finds \(\frac{d^2P}{dx^2}\) of form \(Ax^{-3}\); considers sign or evaluates for positive \(x\) |
| e.g. \(\frac{d^2P}{dx^2}(= 0.4) > 0\) hence minimum (perimeter) | A1 | 2.4 – Requires: \(\frac{d^2P}{dx^2} = 400x^{-3}\); reference to being \(> 0\) for \(x > 0\); correct conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| e.g. \(y = \frac{400 - \pi \times \text{"10"}^2}{8 \times \text{"10"}}\) | M1 | 3.4 – Substitutes their \(x\) value into equation for \(y\), or uses \(P\) and \(x\) to find \(y\) |
| e.g. \(y = 1.07\) (m) so yes this would be suitable | A1 | 2.2a – awrt 1.1 m and concludes suitable. Note "yes suitable because e.g. \(1.07 > 0\)" is A0 |
# Question 12:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Surface area $= 2xy + \frac{\pi x^2}{4}$ | B1 | 1.1b – Correct expression for surface area in terms of $x$ and $y$ only |
| $2xy + \frac{\pi x^2}{4} = 100 \Rightarrow y = \frac{400 - \pi x^2}{8x} = \frac{50}{x} - \frac{\pi x}{8}$ | M1 | 3.4 – Sets expression equal to 100 and rearranges to make $y$ subject |
| $(P =) 2x + 4y + \frac{2\pi x}{4}$ | B1 | 1.1b – Correct expression for perimeter in terms of $x$ and $y$ |
| $(P =) 2x + 4\left(\frac{400 - \pi x^2}{8x}\right) + \frac{2\pi x}{4}$ | M1 | 3.4 – Substitutes their $y$ into perimeter expression |
| $P = 2x + \frac{200}{x}$ * | A1* | 2.1 – cso. Condone omission of $P=$ on final line if seen earlier |
**(5 marks)**
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dP}{dx} = 2 - 200x^{-2}$ | M1, A1 | 1.1b – Attempts differentiation achieving form $A \pm Bx^{-2}$; correct answer |
| $2 - 200x^{-2} = 0 \Rightarrow x = \ldots$ | dM1 | 3.1b – Sets derivative of form $A - Bx^{-2}$ equal to 0, rearranges. Dependent on M1 |
| $x = 10$ | A1 | 1.1b – cao |
**(4 marks)**
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{d^2P}{dx^2} = \text{"400"}x^{-3} \Rightarrow \text{"400"} \times 10^{-3} > 0$ | M1 | 1.1b – Finds $\frac{d^2P}{dx^2}$ of form $Ax^{-3}$; considers sign or evaluates for positive $x$ |
| e.g. $\frac{d^2P}{dx^2}(= 0.4) > 0$ hence minimum (perimeter) | A1 | 2.4 – Requires: $\frac{d^2P}{dx^2} = 400x^{-3}$; reference to being $> 0$ for $x > 0$; correct conclusion |
**(2 marks)**
## Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| e.g. $y = \frac{400 - \pi \times \text{"10"}^2}{8 \times \text{"10"}}$ | M1 | 3.4 – Substitutes their $x$ value into equation for $y$, or uses $P$ and $x$ to find $y$ |
| e.g. $y = 1.07$ (m) so yes this would be suitable | A1 | 2.2a – awrt 1.1 m and concludes suitable. Note "yes suitable because e.g. $1.07 > 0$" is A0 |
**(2 marks)**
**(13 marks total)**
---12.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{23689deb-7eed-4022-848f-1278231a4056-34_494_499_306_778}
\captionsetup{labelformat=empty}
\caption{Figure 5}
\end{center}
\end{figure}
Figure 5 shows the plan view of the design for a swimming pool.\\
The pool is modelled as a quarter of a circle joined to two equal sized rectangles as shown.
Given that
\begin{itemize}
\item the quarter circle has radius $x$ metres
\item the rectangles each have length $x$ metres and width $y$ metres
\item the total surface area of the swimming pool is $100 \mathrm {~m} ^ { 2 }$
\begin{enumerate}[label=(\alph*)]
\item show that, according to the model, the perimeter $P$ metres of the swimming pool is given by
\end{itemize}
$$P = 2 x + \frac { 200 } { x }$$
\item Use calculus to find the value of $x$ for which $P$ has a stationary value.
\item Prove, by further calculus, that this value of $x$ gives a minimum value for $P$
Access to the pool is by side $A B$ shown in Figure 5.\\
Given that $A B$ must be at least one metre,
\item determine, according to the model, whether the swimming pool with the minimum perimeter would be suitable.
\end{enumerate}
\hfill \mbox{\textit{Edexcel AS Paper 1 2024 Q12 [13]}}