| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2018 |
| Session | October |
| Marks | 11 |
| 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 optimization problem requiring area constraint to derive perimeter formula, then differentiation to find minimum. The setup is slightly more complex than basic rectangle problems due to the sector component, but the calculus is routine (differentiate, set to zero, solve). Part (c) requires second derivative test or endpoints check, which is standard A-level technique. Slightly easier than average due to clear structure and straightforward algebraic manipulation. |
| Spec | 1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta1.07n Stationary points: find maxima, minima using derivatives1.07p Points of inflection: using second derivative |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Arc length \(= 0.8x\) | B1 | Correct expression |
| \(P = 2x+4y+0.8x\) | M1 | \(P = \alpha x + \beta y + "0.8x"\), \(\alpha,\beta \neq 0\) |
| \(2xy + \frac{1}{2}(0.8)x^2 = 60\) | B1 | Correct equation for the area |
| \(y = \frac{60-0.4x^2}{2x} \Rightarrow P = 4\left(\frac{60-0.4x^2}{2x}\right)+2.8x\) | M1 | Makes \(y\) the subject and substitutes |
| \(P = \frac{120}{x}+2x\) | A1* | Obtains printed answer with no errors, with \(P=\ldots\) or Perimeter \(=\ldots\) appearing at some point |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dP}{dx} = 2 - \frac{120}{x^2}\) | B1 | Correct derivative |
| \(2 - \frac{120}{x^2} = 0 \Rightarrow x = \sqrt{60}\) | M1 | \(\frac{dP}{dx}=0\) and solves for \(x\). Must be fully correct algebra for their \(\frac{dP}{dx}=0\) which is solvable |
| \(P = \frac{120}{\sqrt{60}}+2\sqrt{60}\) | M1 | Substitutes into \(P\) a positive \(x\) from attempt to solve \(\frac{dP}{dx}=0\) |
| \(P = 4\sqrt{60}\) or \(8\sqrt{15}\) or \(\sqrt{960}\) | A1 | Correct exact answer. cso. Note: if \(\frac{dP}{dx} = 2+\frac{120}{x^2}\) obtained, maximum B0M0M1A0 if positive \(x\) substituted into \(P\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\left(\frac{d^2P}{dx^2}=\right)\frac{240}{x^3} = \frac{240}{(\sqrt{60})^3}\) | M1 | Attempts second derivative (\(x^n \to x^{n-1}\) seen at least once, allow \(k\to 0\) as evidence) and substitutes at least one positive value of \(x\) from \(\frac{dP}{dx}=0\), or makes reference to sign of second derivative provided they have a positive \(x\) |
| \(\left(\frac{d^2P}{dx^2}=\right)\frac{240}{(\sqrt{60})^3} \Rightarrow \frac{d^2P}{dx^2}>0 \therefore\) minimum | A1 | Requires correct second derivative and correct value of \(x\). Must reference sign of second derivative. Allow alternatives e.g. considering values of \(P\) or \(\frac{dP}{dx}\) either side of \(\sqrt{60}\), scoring M1 then A1 if full reason and conclusion given |
## Question 15(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Arc length $= 0.8x$ | B1 | Correct expression |
| $P = 2x+4y+0.8x$ | M1 | $P = \alpha x + \beta y + "0.8x"$, $\alpha,\beta \neq 0$ |
| $2xy + \frac{1}{2}(0.8)x^2 = 60$ | B1 | Correct equation for the area |
| $y = \frac{60-0.4x^2}{2x} \Rightarrow P = 4\left(\frac{60-0.4x^2}{2x}\right)+2.8x$ | M1 | Makes $y$ the subject and substitutes |
| $P = \frac{120}{x}+2x$ | A1* | Obtains printed answer with no errors, with $P=\ldots$ or Perimeter $=\ldots$ appearing at some point |
---
## Question 15(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dP}{dx} = 2 - \frac{120}{x^2}$ | B1 | Correct derivative |
| $2 - \frac{120}{x^2} = 0 \Rightarrow x = \sqrt{60}$ | M1 | $\frac{dP}{dx}=0$ and solves for $x$. Must be fully correct algebra for their $\frac{dP}{dx}=0$ which is solvable |
| $P = \frac{120}{\sqrt{60}}+2\sqrt{60}$ | M1 | Substitutes into $P$ a **positive** $x$ from attempt to solve $\frac{dP}{dx}=0$ |
| $P = 4\sqrt{60}$ or $8\sqrt{15}$ or $\sqrt{960}$ | A1 | Correct exact answer. cso. Note: if $\frac{dP}{dx} = 2+\frac{120}{x^2}$ obtained, maximum B0M0M1A0 if positive $x$ substituted into $P$ |
---
## Question 15(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left(\frac{d^2P}{dx^2}=\right)\frac{240}{x^3} = \frac{240}{(\sqrt{60})^3}$ | M1 | Attempts second derivative ($x^n \to x^{n-1}$ seen at least once, allow $k\to 0$ as evidence) and substitutes at least one **positive** value of $x$ from $\frac{dP}{dx}=0$, **or** makes reference to sign of second derivative provided they have a positive $x$ |
| $\left(\frac{d^2P}{dx^2}=\right)\frac{240}{(\sqrt{60})^3} \Rightarrow \frac{d^2P}{dx^2}>0 \therefore$ minimum | A1 | Requires **correct second derivative** and **correct value of $x$**. Must reference sign of second derivative. Allow alternatives e.g. considering **values** of $P$ or $\frac{dP}{dx}$ either side of $\sqrt{60}$, scoring M1 then A1 if full reason and conclusion given |15.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{1f61f78b-5e77-4758-8ad5-ea00c7dfea2b-46_396_591_251_664}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows a plan for a garden.\\
The garden consists of two identical rectangles of width $y \mathrm {~m}$ and length $x \mathrm {~m}$, joined to a sector of a circle with radius $x \mathrm {~m}$ and angle 0.8 radians, as shown in Figure 2.
The area of the garden is $60 \mathrm {~m} ^ { 2 }$.
\begin{enumerate}[label=(\alph*)]
\item Show that the perimeter, $P \mathrm {~m}$, of the garden is given by
$$P = 2 x + \frac { 120 } { x }$$
\item Use calculus to find the exact minimum value for $P$, giving your answer in the form $a \sqrt { b }$, where $a$ and $b$ are integers.
\item Justify that the value of $P$ found in part (b) is the minimum.
\includegraphics[max width=\textwidth, alt={}, center]{1f61f78b-5e77-4758-8ad5-ea00c7dfea2b-49_83_59_2636_1886}
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2018 Q15 [11]}}