| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2018 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Stationary points and optimisation |
| Type | Open box from cut-corner sheet |
| Difficulty | Moderate -0.5 This is a standard optimization problem from Core Mathematics requiring students to derive a volume formula, differentiate a cubic, solve dV/dx=0, and verify a maximum using second derivative test. While it involves multiple steps (parts a-d), each step follows routine procedures taught in C1/C2 with no novel insight required. The algebra is straightforward and this is a classic textbook-style question, making it slightly easier than average. |
| Spec | 1.02z Models in context: use functions in modelling1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((V =)x(25-2x)(15-2x)\) | M1 | Correct method for the volume. Must be a correct statement for the volume |
| \((V =)x(375 - 80x + 4x^2) = 4x^3 - 80x^2 + 375x\) | A1* | Allow the terms in any order. Completes correctly to printed answer with no errors including bracketing errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dV}{dx} = 12x^2 - 160x + 375\) | M1A1 | M1: \(x^n \rightarrow x^{n-1}\) seen at least once; A1: Correct derivative |
| \(\frac{dV}{dx} = 0 \Rightarrow x = \frac{160 \pm \sqrt{7600}}{24}\) | M1 | Puts \(\frac{dV}{dx} = 0\) (may be implied) and attempts to solve a 3 term quadratic to find \(x\). May be implied by correct values |
| \(x = 3.03,\ 10.3\) but \(0 < x < 7.5\) so \(x = 3.03\) | A1 | Identifies awrt 3.03 only as the required value |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{d^2V}{dx^2} = 24x - 160 = 24(3.03) - 160\) | M1 | Attempts the second derivative \((x^n \rightarrow x^{n-1})\) and substitutes at least one positive value of \(x\) from their \(\frac{dV}{dx} = 0\) |
| \(\frac{d^2V}{dx^2} = 24(3.03) - 160 \Rightarrow \frac{d^2V}{dx^2} < 0\ \therefore\) maximum | A1 | Fully correct proof using correct second derivative and \(x =\) awrt 3 only. Must be substitution and reference to sign of second derivative. Accept "negative so \(x\) is the maximum" |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(V = 4(3.03)^3 - 80(3.03)^2 + 375(3.03)\) | M1 | Substitutes a (positive) \(x\) from their \(\frac{dV}{dx} = 0\) into the given \(V\) or a "version" of \(V\) |
| \(V = 513\) | A1 | Awrt 513 |
# Question 7:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(V =)x(25-2x)(15-2x)$ | M1 | Correct method for the volume. Must be a correct statement for the volume |
| $(V =)x(375 - 80x + 4x^2) = 4x^3 - 80x^2 + 375x$ | A1* | Allow the terms in any order. Completes correctly to printed answer with no errors including bracketing errors |
**(2 marks)**
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dV}{dx} = 12x^2 - 160x + 375$ | M1A1 | M1: $x^n \rightarrow x^{n-1}$ seen at least once; A1: Correct derivative |
| $\frac{dV}{dx} = 0 \Rightarrow x = \frac{160 \pm \sqrt{7600}}{24}$ | M1 | Puts $\frac{dV}{dx} = 0$ (may be implied) and attempts to solve a 3 term quadratic to find $x$. May be implied by correct values |
| $x = 3.03,\ 10.3$ but $0 < x < 7.5$ so $x = 3.03$ | A1 | Identifies awrt 3.03 only as the required value |
**(4 marks)**
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{d^2V}{dx^2} = 24x - 160 = 24(3.03) - 160$ | M1 | Attempts the second derivative $(x^n \rightarrow x^{n-1})$ and substitutes at least one **positive** value of $x$ from their $\frac{dV}{dx} = 0$ |
| $\frac{d^2V}{dx^2} = 24(3.03) - 160 \Rightarrow \frac{d^2V}{dx^2} < 0\ \therefore$ maximum | A1 | Fully correct proof using correct second derivative and $x =$ awrt 3 only. Must be substitution and reference to sign of second derivative. Accept "negative so $x$ is the maximum" |
**(2 marks)**
## Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $V = 4(3.03)^3 - 80(3.03)^2 + 375(3.03)$ | M1 | Substitutes a (positive) $x$ from their $\frac{dV}{dx} = 0$ into the given $V$ or a "version" of $V$ |
| $V = 513$ | A1 | Awrt 513 |
**(2 marks) — Total 10**
---7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{6f9ace43-747b-419f-a9d1-d30165d77379-18_675_1408_292_358}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a rectangular sheet of metal of negligible thickness, which measures 25 cm by 15 cm . Squares of side $x \mathrm {~cm}$ are cut from each corner of the sheet and the remainder is folded along the dotted lines to make an open cuboid box, as shown in Figure 2.
\begin{enumerate}[label=(\alph*)]
\item Show that the volume, $V \mathrm {~cm} ^ { 3 }$, of the box is given by
$$V = 4 x ^ { 3 } - 80 x ^ { 2 } + 375 x$$
\item Use calculus to find the value of $x$, to 3 significant figures, for which the volume of the box is a maximum.
\item Justify that this value of $x$ gives a maximum value for $V$.
\item Find, to 3 significant figures, the maximum volume of the box.
\begin{center}
\end{center}
\section*{8.}
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{6f9ace43-747b-419f-a9d1-d30165d77379-22_670_1004_292_392}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Figure 3 shows a sketch of the curve with equation $y = \mathrm { f } ( x ) , x \in \mathbb { R }$.
The curve crosses the $y$-axis at the point $( 0,5 )$ and crosses the $x$-axis at the point $( 6,0 )$.
The curve has a minimum point at $( 1,3 )$ and a maximum point at $( 4,7 )$.
On separate diagrams, sketch the curve with equation
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2018 Q7 [10]}}