| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2014 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Stationary points and optimisation |
| Type | Show formula then optimise: cylinder/prism (single variable) |
| Difficulty | Standard +0.3 This is a standard C2 optimization problem with guided steps. Part (a) uses volume formula with given trapezium area, (b) requires systematic surface area calculation, (c) is routine differentiation and solving dS/dx=0, and (d) applies the second derivative test. All steps are clearly signposted 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 | Marks | Guidance |
| \(\frac{1}{2}(9x+6x)4x\) or \(2x \times 15x\) or \(\left(\frac{1}{2}4x\times(9x-6x)+6x\times4x\right)\) or \(6x^2+24x^2\) or \(\left(9x\times4x-\frac{1}{2}4x\times(9x-6x)\right)\) or \(36x^2-6x^2\) | M1 | Correct attempt at area of a trapezium. Note \(30x^2\) on its own or \(30x^2\) from incorrect work e.g. \(5x\times6x\) is M0. If clear intention to find trapezium area correctly allow M1 but A1 can be withheld if there are slips. |
| \(\Rightarrow 30x^2y = 9600 \Rightarrow y = \frac{9600}{30x^2} \Rightarrow y = \frac{320}{x^2}\) * | A1 cso | Correct proof with at least one intermediate step and no errors seen. "\(y=\)" is required. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((S=)\frac{1}{2}(9x+6x)4x + \frac{1}{2}(9x+6x)4x + 6xy + 9xy + 5xy + 4xy\) | M1 | Attempt to find area of six faces. The 2 trapezia may be combined as \((9x+6x)4x\) or \(60x^2\) and the 4 other faces combined as \(24xy\) but all six faces must be included. Must attempt areas of two trapezia that are dimensionally correct. |
| Correct expression in any form. Allow just \((S=)60x^2+24xy\) | A1 | Correct expression in any form. |
| \(y=\frac{320}{x^2} \Rightarrow (S=)30x^2+30x^2+24x\left(\frac{320}{x^2}\right)\) | M1 | Substitutes \(y=\frac{320}{x^2}\) into their expression for \(S\). \(S\) should have at least one \(x^2\) term and one \(xy\) term but there may be other terms which may be dimensionally incorrect. |
| So, \((S=)60x^2+\frac{7680}{x}\) * | A1* cso | Correct solution only. "\(S=\)" is not required here. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dS}{dx}=120x-7680x^{-2}\) \(\left\{=120x-\frac{7680}{x^2}\right\}\) | M1 | Either \(60x^2\rightarrow120x\) or \(\frac{7680}{x}\rightarrow\frac{\pm\lambda}{x^2}\) |
| Correct differentiation (need not be simplified) | A1 aef | |
| \(120x - \frac{7680}{x^2}=0\) \(\Rightarrow x^3=\frac{7680}{120}=64 \Rightarrow x=4\) | M1 | \(S'=0\) and "their \(x^3=\pm\) value" or "their \(x^{-3}=\pm\) value". Setting \(\frac{dS}{dx}=0\) and candidate's first correct power of \(x\) = a value. Power of \(x\) must be consistent with their differentiation. \(S'=0\) can be implied by \(120x=\frac{7680}{x^2}\). \(x=4\) only (\(x^3=64\Rightarrow x=\pm4\) scores A0) |
| \(x=4\) only | A1 cso | |
| \(\{x=4\}\), \(S=60(4)^2+\frac{7680}{4}=2880\) (cm²) | ddM1 | Substitute candidate's value of \(x\ (\neq0)\) into formula for \(S\). Dependent on both previous M marks. |
| \(2880\) | A1 cao and cso | 2880 cso (Must come from correct work) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{d^2S}{dx^2}=120+\frac{15360}{x^3}>0 \Rightarrow\) Minimum | M1 | Attempt \(S''\ (x^n\rightarrow x^{n-1})\) and considers sign. Requires attempt at second derivative and some consideration of its sign. An attempt to solve \(S''=0\) is M0. |
| \(120+\frac{15360}{x^3}\) and \(>0\) and conclusion | A1 ft | Requires correct second derivative of \(120+\frac{15360}{x^3}\) (need not be simplified) and valid reason (e.g. \(>0\)) and conclusion. Follow through only through a correct second derivative i.e. \(x\) may be incorrect but must be positive and/or \(S''\) may have been evaluated incorrectly. |
# Question 10:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{2}(9x+6x)4x$ or $2x \times 15x$ or $\left(\frac{1}{2}4x\times(9x-6x)+6x\times4x\right)$ or $6x^2+24x^2$ or $\left(9x\times4x-\frac{1}{2}4x\times(9x-6x)\right)$ or $36x^2-6x^2$ | M1 | Correct attempt at area of a trapezium. Note $30x^2$ on its own or $30x^2$ from incorrect work e.g. $5x\times6x$ is M0. If clear intention to find trapezium area correctly allow M1 but A1 can be withheld if there are slips. |
| $\Rightarrow 30x^2y = 9600 \Rightarrow y = \frac{9600}{30x^2} \Rightarrow y = \frac{320}{x^2}$ * | A1 cso | Correct proof with at least one intermediate step and no errors seen. **"$y=$" is required.** |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(S=)\frac{1}{2}(9x+6x)4x + \frac{1}{2}(9x+6x)4x + 6xy + 9xy + 5xy + 4xy$ | M1 | Attempt to find area of **six** faces. The 2 trapezia may be combined as $(9x+6x)4x$ or $60x^2$ and the 4 other faces combined as $24xy$ but all six faces must be included. Must attempt areas of two trapezia that are dimensionally correct. |
| Correct expression in any form. Allow just $(S=)60x^2+24xy$ | A1 | Correct expression in any form. |
| $y=\frac{320}{x^2} \Rightarrow (S=)30x^2+30x^2+24x\left(\frac{320}{x^2}\right)$ | M1 | Substitutes $y=\frac{320}{x^2}$ into their expression for $S$. $S$ should have at least one $x^2$ term and one $xy$ term but there may be other terms which may be dimensionally incorrect. |
| So, $(S=)60x^2+\frac{7680}{x}$ * | A1* cso | Correct solution only. "$S=$" is **not** required here. |
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dS}{dx}=120x-7680x^{-2}$ $\left\{=120x-\frac{7680}{x^2}\right\}$ | M1 | Either $60x^2\rightarrow120x$ or $\frac{7680}{x}\rightarrow\frac{\pm\lambda}{x^2}$ |
| Correct differentiation (need not be simplified) | A1 aef | |
| $120x - \frac{7680}{x^2}=0$ $\Rightarrow x^3=\frac{7680}{120}=64 \Rightarrow x=4$ | M1 | $S'=0$ and "their $x^3=\pm$ value" or "their $x^{-3}=\pm$ value". Setting $\frac{dS}{dx}=0$ and candidate's first **correct** power of $x$ = a value. **Power of $x$ must be consistent with their differentiation.** $S'=0$ can be implied by $120x=\frac{7680}{x^2}$. $x=4$ only ($x^3=64\Rightarrow x=\pm4$ scores A0) |
| $x=4$ only | A1 cso | |
| $\{x=4\}$, $S=60(4)^2+\frac{7680}{4}=2880$ (cm²) | ddM1 | Substitute candidate's value of $x\ (\neq0)$ into formula for $S$. **Dependent on both previous M marks.** |
| $2880$ | A1 cao and cso | 2880 cso (Must come from correct work) |
## Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{d^2S}{dx^2}=120+\frac{15360}{x^3}>0 \Rightarrow$ Minimum | M1 | Attempt $S''\ (x^n\rightarrow x^{n-1})$ **and** considers sign. Requires attempt at second derivative and **some consideration of its sign**. An attempt to solve $S''=0$ is M0. |
| $120+\frac{15360}{x^3}$ and $>0$ **and** conclusion | A1 ft | Requires **correct** second derivative of $120+\frac{15360}{x^3}$ (need not be simplified) **and** valid reason (e.g. $>0$) **and** conclusion. Follow through only through a correct second derivative i.e. $x$ may be incorrect **but must be positive** and/or $S''$ may have been evaluated incorrectly. |
> **Note:** A correct $S''$ followed by $S''("4")="360"$ therefore minimum scores **no marks** in (d). A correct $S''$ followed by $S''("4")="360"$ which is positive therefore minimum scores **both marks**.
> **Note:** Parts (c) and (d) can be marked together. **Total: 14 marks**10.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{e6b490c0-80c4-4e15-b587-ac052ee27db7-17_929_584_237_287}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{e6b490c0-80c4-4e15-b587-ac052ee27db7-17_716_544_452_1069}
\captionsetup{labelformat=empty}
\caption{Figure 5}
\end{center}
\end{figure}
Figure 4 shows a closed letter box $A B F E H G C D$, which is made to be attached to a wall of a house.
The letter box is a right prism of length $y \mathrm {~cm}$ as shown in Figure 4. The base $A B F E$ of the prism is a rectangle. The total surface area of the six faces of the prism is $S \mathrm {~cm} ^ { 2 }$.
The cross section $A B C D$ of the letter box is a trapezium with edges of lengths $D A = 9 x \mathrm {~cm}$, $A B = 4 x \mathrm {~cm} , B C = 6 x \mathrm {~cm}$ and $C D = 5 x \mathrm {~cm}$ as shown in Figure 5.\\
The angle $D A B = 90 ^ { \circ }$ and the angle $A B C = 90 ^ { \circ }$.
The volume of the letter box is $9600 \mathrm {~cm} ^ { 3 }$.
\begin{enumerate}[label=(\alph*)]
\item Show that
$$y = \frac { 320 } { x ^ { 2 } }$$
\item Hence show that the surface area of the letter box, $S \mathrm {~cm} ^ { 2 }$, is given by
$$S = 60 x ^ { 2 } + \frac { 7680 } { x }$$
\item Use calculus to find the minimum value of $S$.
\item Justify, by further differentiation, that the value of $S$ you have found is a minimum.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2014 Q10 [14]}}