| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2019 |
| Session | January |
| Marks | 16 |
| 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 optimization problem requiring volume constraint to derive surface area formula, then differentiation to minimize. All steps are routine A-level techniques (Pythagoras, basic calculus, second derivative test) with clear scaffolding across parts (a)-(d). Slightly easier than average due to heavy guidance and straightforward algebra. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx |
| END |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{1}{2} \times 4x \times 3x \times l = 0.75 \Rightarrow l = \frac{1}{8x^2}\) | M1 A1 | Uses \(kx^2l = 0.75 \Rightarrow l = ..\) or \(kx^2l = 0.75 \Rightarrow lx = ..\) giving correct dimensions. A1 for \(l = \frac{1}{8x^2}\) or equivalent e.g. \(lx = \frac{1}{8x}\), allow \(l = \frac{1.5}{12x^2}\) |
| States or uses \(S = 3xl + 4xl + 6x^2 + 6x^2\) | B1 | Accept \((S=)3xl + 4xl + 6x^2 + 6x^2\) or exact equivalent. Accept \((S=)7xl + 12x^2\). Condone lack of \(S=\) and allow other letters. Allow unsimplified e.g. \(\frac{1}{2}\times 4x \times 3x\) |
| Substitute \(l = \frac{1}{8x^2}\) in \(S = 12x^2 + 7xl \Rightarrow 12x^2 + 7x \times \frac{1}{8x^2} \Rightarrow S = 12x^2 + \frac{7}{8x}\) | M1, A1* | M1: substitutes their \(l=...\) into \(S = Cxl + Dx^2\). A1*: given answer, all calculations must be correct, no incorrect lines, LHS \(S=\) or \(SA=\) or Surface Area \(=\) must appear somewhere |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\left(\frac{dS}{dx}\right) = 24x - \frac{7}{8x^2}\) | M1A1 | M1 for \(\left(\frac{dS}{dx}\right) = Px + Qx^{-2}\). A1 for \(24x - \frac{7}{8x^2}\) which may be unsimplified |
| Sets \(24x - \frac{7}{8x^2} = 0 \Rightarrow x^3 = \frac{7}{192}\) | M1 A1 | M1: sets their \(24x - \frac{7}{8x^2} = 0\) and proceeds to \(x^n = C\), \(C>0\) or \(\frac{1}{x^n} = D\), \(D>0\) with \(n \neq 1\). A1: \(x^3 = \frac{7}{192}\) or \(x^{-3} = \frac{192}{7}\), accept awrt \(x^3 = 0.036\) |
| \(\Rightarrow x = 0.332\) (m) cao | A1 | This answer only here (following correct work). Correct answer following correct differentiation implies final three marks |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\left(\frac{d^2S}{dx^2}\right) = 24 + \frac{7}{4x^3}\) | M1 | Score for: achieving \(\frac{d^2S}{dx^2} = A + \frac{B}{x^3}\) and attempting to find its value at their positive \(x=0.332\); OR achieving \(\frac{d^2S}{dx^2} = A + \frac{B}{x^3}\) and attempting to justify \(\frac{d^2S}{dx^2} > 0\) as \(x>0\); OR achieving \(\frac{dS}{dx} = Cx + \frac{D}{x^2}\) and attempting sign/value either side of \(x=0.332\); OR finding \(S\) at \(x=0.332\) and at \(x=0.331\) and \(x=0.333\) |
| When \(x = 0.332\) m, \(\frac{d^2S}{dx^2} > 0\) hence minimum | A1 | Requires correct expression for \(\frac{d^2S}{dx^2}\), correct value for \(x\) (awrt 0.33), correct \(\frac{d^2S}{dx^2}\) value, and acceptable conclusion. Values correct to 1 sf. e.g. \(\left(\frac{d^2S}{dx^2}\right) = 24 + \frac{7}{4\times0.332^3}=\) awrt \(70 > 0\). Accept 'it is a minimum' but not 'hence \(x\) is a minimum' |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Length \(AD = 5x = 5 \times \text{"0.332"} =\) awrt \(1.66\) (m) | M1A1 | M1: attempts \(5x = 5\times\text{"0.332"}\) where "0.332" is their answer to (b). Alternatively may use Pythagoras with \(4\times\text{"0.332"}\) and \(3\times\text{"0.332"}\). Condone \(AD^2 = 25x^2 \Rightarrow AD = 25x\) as attempt at square root. A1: awrt 1.66 (metres) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Length \(CD = l = \frac{1}{8(\text{"0.332"})^2} =\) awrt \(1.13/1.14\) (m) | M1 A1 | M1: attempts value of \(CD\) using their \(l = \frac{\alpha}{\beta x^2}\) or equivalent, ft on their \(x\). Note possible to find \(l\) using \(x=\text{"0.332"}\) and \(S_{0.332}\) in \(S = Cxl + Dx^2\). A1: awrt 1.13/1.14 (metres) |
# Question 16:
## Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{1}{2} \times 4x \times 3x \times l = 0.75 \Rightarrow l = \frac{1}{8x^2}$ | M1 A1 | Uses $kx^2l = 0.75 \Rightarrow l = ..$ or $kx^2l = 0.75 \Rightarrow lx = ..$ giving correct dimensions. A1 for $l = \frac{1}{8x^2}$ or equivalent e.g. $lx = \frac{1}{8x}$, allow $l = \frac{1.5}{12x^2}$ |
| States or uses $S = 3xl + 4xl + 6x^2 + 6x^2$ | B1 | Accept $(S=)3xl + 4xl + 6x^2 + 6x^2$ or exact equivalent. Accept $(S=)7xl + 12x^2$. Condone lack of $S=$ and allow other letters. Allow unsimplified e.g. $\frac{1}{2}\times 4x \times 3x$ |
| Substitute $l = \frac{1}{8x^2}$ in $S = 12x^2 + 7xl \Rightarrow 12x^2 + 7x \times \frac{1}{8x^2} \Rightarrow S = 12x^2 + \frac{7}{8x}$ | M1, A1* | M1: substitutes their $l=...$ into $S = Cxl + Dx^2$. A1*: given answer, all calculations must be correct, no incorrect lines, LHS $S=$ or $SA=$ or Surface Area $=$ must appear somewhere |
**(5 marks)**
## Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\left(\frac{dS}{dx}\right) = 24x - \frac{7}{8x^2}$ | M1A1 | M1 for $\left(\frac{dS}{dx}\right) = Px + Qx^{-2}$. A1 for $24x - \frac{7}{8x^2}$ which may be unsimplified |
| Sets $24x - \frac{7}{8x^2} = 0 \Rightarrow x^3 = \frac{7}{192}$ | M1 A1 | M1: sets their $24x - \frac{7}{8x^2} = 0$ and proceeds to $x^n = C$, $C>0$ or $\frac{1}{x^n} = D$, $D>0$ with $n \neq 1$. A1: $x^3 = \frac{7}{192}$ or $x^{-3} = \frac{192}{7}$, accept awrt $x^3 = 0.036$ |
| $\Rightarrow x = 0.332$ (m) cao | A1 | This answer only here (following correct work). Correct answer following correct differentiation implies final three marks |
**(5 marks)**
## Part (c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\left(\frac{d^2S}{dx^2}\right) = 24 + \frac{7}{4x^3}$ | M1 | Score for: achieving $\frac{d^2S}{dx^2} = A + \frac{B}{x^3}$ and attempting to find its value at their positive $x=0.332$; OR achieving $\frac{d^2S}{dx^2} = A + \frac{B}{x^3}$ and attempting to justify $\frac{d^2S}{dx^2} > 0$ as $x>0$; OR achieving $\frac{dS}{dx} = Cx + \frac{D}{x^2}$ and attempting sign/value either side of $x=0.332$; OR finding $S$ at $x=0.332$ and at $x=0.331$ and $x=0.333$ |
| When $x = 0.332$ m, $\frac{d^2S}{dx^2} > 0$ hence minimum | A1 | Requires correct expression for $\frac{d^2S}{dx^2}$, correct value for $x$ (awrt 0.33), correct $\frac{d^2S}{dx^2}$ value, and acceptable conclusion. Values correct to 1 sf. e.g. $\left(\frac{d^2S}{dx^2}\right) = 24 + \frac{7}{4\times0.332^3}=$ awrt $70 > 0$. Accept 'it is a minimum' but not 'hence $x$ is a minimum' |
**(2 marks)**
## Part (d)(i):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Length $AD = 5x = 5 \times \text{"0.332"} =$ awrt $1.66$ (m) | M1A1 | M1: attempts $5x = 5\times\text{"0.332"}$ where "0.332" is their answer to (b). Alternatively may use Pythagoras with $4\times\text{"0.332"}$ and $3\times\text{"0.332"}$. Condone $AD^2 = 25x^2 \Rightarrow AD = 25x$ as attempt at square root. A1: awrt 1.66 (metres) |
## Part (d)(ii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Length $CD = l = \frac{1}{8(\text{"0.332"})^2} =$ awrt $1.13/1.14$ (m) | M1 A1 | M1: attempts value of $CD$ using their $l = \frac{\alpha}{\beta x^2}$ or equivalent, ft on their $x$. Note possible to find $l$ using $x=\text{"0.332"}$ and $S_{0.332}$ in $S = Cxl + Dx^2$. A1: awrt 1.13/1.14 (metres) |
**(4 marks total for d)**
**(16 marks total)**16.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{75d68987-2314-4c8f-8160-24977c5c4e34-44_442_822_285_561}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
Figure 4 shows the design for a container in the shape of a hollow triangular prism.
The container is open at the top, which is labelled $A B C D$.
The sides of the container, $A B F E$ and $D C F E$, are rectangles.
The ends of the container, $A D E$ and $B C F$, are congruent right-angled triangles, as shown in Figure 4.
The ends of the container are vertical and the edge $E F$ is horizontal.
The edges $A E , D E$ and $E F$ have lengths $4 x$ metres, $3 x$ metres and $l$ metres respectively.
Given that the container has a capacity of $0.75 \mathrm {~m} ^ { 3 }$ and is made of material of negligible thickness,
\begin{enumerate}[label=(\alph*)]
\item show that the internal surface area of the container, $S \mathrm {~m} ^ { 2 }$, is given by
$$S = 12 x ^ { 2 } + \frac { 7 } { 8 x }$$
\item Use calculus to find the value of $x$, for which $S$ is a minimum.
Give your answer to 3 significant figures.
\item Justify that the value of $x$ found in part (b) gives a minimum value for $S$.
Using the value of $x$ found in part (b), find to 2 decimal places,
\item \begin{enumerate}[label=(\roman*)]
\item the length of the edge $A D$,
\item the length of the edge $C D$.
\begin{center}
\begin{tabular}{|l|l|}
\hline
\hline
END & \\
\hline
\end{tabular}
\end{center}
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2019 Q16 [16]}}