| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2019 |
| Session | June |
| Marks | 11 |
| 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 constraint formulation from area, differentiation to find stationary points, and second derivative test. While multi-part (4 parts, likely 8-10 marks), each step follows a predictable routine: derive P(x) from given area constraint, set dP/dx=0, verify minimum, evaluate. The algebra is straightforward and the problem type is extremely common in C1/C2 textbooks. Slightly easier than average due to its formulaic nature. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Area \(= xy + \frac{1}{2}\pi\left(\frac{x}{2}\right)^2\) or \(P = 2y + x + \frac{1}{2}\pi x\) oe | B1 | |
| Both Area and \(P\) expressions correct | B1 | |
| \(y = \frac{100}{x} - \frac{\pi x}{8} \Rightarrow P = 2\left(\frac{100}{x}-\frac{\pi x}{8}\right)+x+\frac{\pi x}{2}\) | M1 | |
| \(P = \frac{1}{4}x(4+\pi)+\frac{200}{x}\) * | A1* | Printed answer, must be shown |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\left(\frac{dP}{dx} =\right)\ 1+\frac{\pi}{4}-\frac{200}{x^2}\) | B1 | |
| \(1+\frac{\pi}{4}-\frac{200}{x^2}=0 \Rightarrow x^2 = \frac{800}{4+\pi}\) | M1 | |
| \(x = \sqrt{\frac{800}{4+\pi}}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\left(\frac{d^2P}{dx^2}=\right)\frac{400}{x^3} = \frac{400}{\sqrt{\frac{800}{4+\pi}}}=\ldots\) or makes reference to sign of \(\frac{d^2P}{dx^2}\) | M1 | Alt: substitute \(x=10, x=11\) into \(\frac{dP}{dx}\) to show sign change |
| As \(\frac{d^2P}{dx^2} = \frac{400}{x^3} > 0\) when \(x>0\), \(P\) is a minimum | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P = \frac{1}{4}x(4+\pi)+\frac{200}{x} = \ldots\) | M1 | Substitutes their \(x\) value |
| \(= \text{awrt } 37.8\ \text{(m)}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(xy + \frac{1}{2}\pi\left(\frac{x}{2}\right)^2\) seen or implied | B1 | oe OR \(2y + x + \frac{1}{2}\pi x\) oe, possibly unsimplified |
| Sight or implied use of both \(2y + x + \frac{1}{2}\pi x\) AND \(xy + \frac{1}{2}\pi\left(\frac{x}{2}\right)^2\) | B1 | oe, possibly unsimplified |
| Sets area = 100, rearranges for \(y\), substitutes into perimeter expression | M1 | Don't be concerned with mechanics of rearranging |
| Achieves given answer with no errors including brackets | A1* | cso |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dP}{dx} = 1 + \frac{\pi}{4} - \frac{200}{x^2}\) | B1 | oe, unsimplified although power must be processed |
| Sets \(\frac{dP}{dx} = 0\), must be of form \(\frac{dP}{dx} = A + Bx^{-2}\), \(A\neq 0\), \(B < 0\), proceeds to \(x^{+2} = C\) where \(C > 0\) | M1 | awrt 112 or awrt 0.00893 sufficient evidence |
| \(x = \sqrt{\frac{800}{4+\pi}}\) or exact equivalent e.g. \(\frac{20\sqrt{2}}{\sqrt{4+\pi}}\) | A1 | Do not accept decimals; no fractions on numerator or denominator; \(\pm\sqrt{\frac{800}{4+\pi}}\) is A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Finds \(\frac{d^2P}{dx^2}\) of form \(Ax^{-3}\), \(A > 0\), and either substitutes \(x\) value or refers to sign | M1 | Alternatively substitute \(x\) values either side into \(\frac{dP}{dx}\) |
| \(\frac{d^2P}{dx^2} = \frac{400}{x^3}\), value of \(x = 10.6\) or better, concludes minimum since \(\frac{400}{x^3} > 0\) when \(x > 0\) | A1 | cso; calculated value must be correct (0.3, 1sf); alternative method values must be correct (1sf) with gradient change from negative to positive |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Substitutes \(x\) value from (b) into given equation for \(P\) | M1 | Sufficient to see value substituted into expression followed by answer |
| awrt 37.8 | A1 |
# Question 15:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Area $= xy + \frac{1}{2}\pi\left(\frac{x}{2}\right)^2$ or $P = 2y + x + \frac{1}{2}\pi x$ oe | B1 | |
| Both Area and $P$ expressions correct | B1 | |
| $y = \frac{100}{x} - \frac{\pi x}{8} \Rightarrow P = 2\left(\frac{100}{x}-\frac{\pi x}{8}\right)+x+\frac{\pi x}{2}$ | M1 | |
| $P = \frac{1}{4}x(4+\pi)+\frac{200}{x}$ * | A1* | Printed answer, must be shown |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left(\frac{dP}{dx} =\right)\ 1+\frac{\pi}{4}-\frac{200}{x^2}$ | B1 | |
| $1+\frac{\pi}{4}-\frac{200}{x^2}=0 \Rightarrow x^2 = \frac{800}{4+\pi}$ | M1 | |
| $x = \sqrt{\frac{800}{4+\pi}}$ | A1 | |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left(\frac{d^2P}{dx^2}=\right)\frac{400}{x^3} = \frac{400}{\sqrt{\frac{800}{4+\pi}}}=\ldots$ or makes reference to sign of $\frac{d^2P}{dx^2}$ | M1 | Alt: substitute $x=10, x=11$ into $\frac{dP}{dx}$ to show sign change |
| As $\frac{d^2P}{dx^2} = \frac{400}{x^3} > 0$ when $x>0$, $P$ is a minimum | A1 | |
## Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P = \frac{1}{4}x(4+\pi)+\frac{200}{x} = \ldots$ | M1 | Substitutes their $x$ value |
| $= \text{awrt } 37.8\ \text{(m)}$ | A1 | |
# Question (Garden Optimization Problem):
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $xy + \frac{1}{2}\pi\left(\frac{x}{2}\right)^2$ seen or implied | B1 | oe OR $2y + x + \frac{1}{2}\pi x$ oe, possibly unsimplified |
| Sight or implied use of both $2y + x + \frac{1}{2}\pi x$ AND $xy + \frac{1}{2}\pi\left(\frac{x}{2}\right)^2$ | B1 | oe, possibly unsimplified |
| Sets area = 100, rearranges for $y$, substitutes into perimeter expression | M1 | Don't be concerned with mechanics of rearranging |
| Achieves given answer with no errors including brackets | A1* | cso |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dP}{dx} = 1 + \frac{\pi}{4} - \frac{200}{x^2}$ | B1 | oe, unsimplified although power must be processed |
| Sets $\frac{dP}{dx} = 0$, must be of form $\frac{dP}{dx} = A + Bx^{-2}$, $A\neq 0$, $B < 0$, proceeds to $x^{+2} = C$ where $C > 0$ | M1 | awrt 112 or awrt 0.00893 sufficient evidence |
| $x = \sqrt{\frac{800}{4+\pi}}$ or exact equivalent e.g. $\frac{20\sqrt{2}}{\sqrt{4+\pi}}$ | A1 | Do not accept decimals; no fractions on numerator or denominator; $\pm\sqrt{\frac{800}{4+\pi}}$ is A0 |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Finds $\frac{d^2P}{dx^2}$ of form $Ax^{-3}$, $A > 0$, and either substitutes $x$ value or refers to sign | M1 | Alternatively substitute $x$ values either side into $\frac{dP}{dx}$ |
| $\frac{d^2P}{dx^2} = \frac{400}{x^3}$, value of $x = 10.6$ or better, concludes minimum since $\frac{400}{x^3} > 0$ when $x > 0$ | A1 | cso; calculated value must be correct (0.3, 1sf); alternative method values must be correct (1sf) with gradient change from negative to positive |
## Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Substitutes $x$ value from (b) into given equation for $P$ | M1 | Sufficient to see value substituted into expression followed by answer |
| awrt 37.8 | A1 | |
---15.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{de511cb3-35c7-4225-b459-a136b6304b78-44_537_679_258_589}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Diagram not drawn to scale
Figure 3 shows the plan view of a garden. The shape of this garden consists of a rectangle joined to a semicircle.
The rectangle has length $x$ metres and width $y$ metres.\\
The area of the garden is $100 \mathrm {~m} ^ { 2 }$.
\begin{enumerate}[label=(\alph*)]
\item Show that the perimeter, $P$ metres, of the garden is given by
$$P = \frac { 1 } { 4 } x ( 4 + \pi ) + \frac { 200 } { x } \quad x > 0$$
\item Use calculus to find the exact value of $x$ for which the perimeter of the garden is a minimum.
\item Justify that the value of $x$ found in part (b) gives a minimum value for $P$.
\item Find the minimum perimeter of the garden, giving your answer in metres to one decimal place.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2019 Q15 [11]}}