| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Optimization with constraints |
| Difficulty | Standard +0.3 This is a standard C2 optimization problem involving forming an expression for volume, finding the domain, and using differentiation to find a maximum. Part (a) is algebraic manipulation, parts (c)-(e) are routine calculus (differentiate, solve dV/dx=0, substitute back, verify with second derivative). While multi-part with 12 marks total, each step follows a well-practiced procedure with no novel insight required, making it slightly easier than the average A-level question. |
| 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/dx1.07p Points of inflection: using second derivative |
| Answer | Marks | Guidance |
|---|---|---|
| \(l = (50 - 2x)\), \(w = (40 - 2x)\) | B1 | |
| \(V = x(50 - 2x)(40 - 2x)\) with \(V = x \cdot l \cdot w\) | M1 | |
| \(V = x(2000 - 80x - 100x + 4x^2) = 4x(x^2 - 45x + 500)\) | (*) | A1 cso |
| Answer | Marks | Guidance |
|---|---|---|
| \(0 < x < 20\) (accept \(\le\)) | B1 | (1 mark) |
| Answer | Marks |
|---|---|
| \(\frac{dV}{dx} = 12x^2 - 360x + 2000\) (accept \(\div\) 4) | M1, A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dV}{dx} = 0 \Rightarrow 3x^2 - 90x + 500 = 0 \Rightarrow x = \frac{90 \pm \sqrt{8100 - 6000}}{6}\) | M1 (\(dV/dx = 0\) & attempt to solve) | |
| \(x = (22.6)\), required \(x = 7.36\) or \(7.4\) or \(7.362\) | A1 | (4 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(V_{\max} = 4 \times 7.36(7.36^2\ldots) = 6564\) or \(6560\) or \(6600\) | M1, A1 | (2 marks) |
| Or: e.g. \(V'' = 24x - 360 | _{x=7.36} = (-183\ldots) < 0\), \(\therefore\) maximum | M1 full method, A1 full accuracy |
**Part (a)**
$l = (50 - 2x)$, $w = (40 - 2x)$ | B1 |
$V = x(50 - 2x)(40 - 2x)$ with $V = x \cdot l \cdot w$ | M1 |
$V = x(2000 - 80x - 100x + 4x^2) = 4x(x^2 - 45x + 500)$ | (*) | A1 cso | (3 marks)
**Part (b)**
$0 < x < 20$ (accept $\le$) | B1 | (1 mark)
**Part (c)**
$\frac{dV}{dx} = 12x^2 - 360x + 2000$ (accept $\div$ 4) | M1, A1 |
**Part (d)**
$\frac{dV}{dx} = 0 \Rightarrow 3x^2 - 90x + 500 = 0 \Rightarrow x = \frac{90 \pm \sqrt{8100 - 6000}}{6}$ | M1 ($dV/dx = 0$ & attempt to solve) |
$x = (22.6)$, required $x = 7.36$ or $7.4$ or $7.362$ | A1 | (4 marks)
**Part (e)**
$V_{\max} = 4 \times 7.36(7.36^2\ldots) = 6564$ or $6560$ or $6600$ | M1, A1 | (2 marks)
Or: e.g. $V'' = 24x - 360|_{x=7.36} = (-183\ldots) < 0$, $\therefore$ maximum | M1 full method, A1 full accuracy | (2 marks)
**Total for Question 9: 12 marks**\includegraphics{figure_3}
A rectangular sheet of metal measures 50 cm by 40 cm. Squares of side $x$ cm are cut from each corner of the sheet and the remainder is folded along the dotted lines to make an open tray, as shown in Fig. 3.
\begin{enumerate}[label=(\alph*)]
\item Show that the volume, $V$ cm$^3$, of the tray is given by $V = 4x(x^2 - 45x + 500)$. [3]
\item State the range of possible values of $x$. [1]
\item Find the value of $x$ for which $V$ is a maximum. [4]
\item Hence find the maximum value of $V$. [2]
\item Justify that the value of $V$ you found in part (d) is a maximum. [2]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 Q9 [12]}}