| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Stationary points and optimisation |
| Type | Show formula then optimise: composite/irregular shape |
| Difficulty | Challenging +1.2 This is a multi-part optimization problem requiring volume calculation from a complex 3D shape, surface area formulation, and standard calculus techniques (differentiation, finding stationary points, second derivative test). While it involves several steps and careful geometric reasoning to set up the area formula, the calculus itself is routine C2 level. The geometric complexity and algebraic manipulation elevate it above average difficulty, but it remains a standard textbook-style optimization problem without requiring novel insight. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx |
| Answer | Marks |
|---|---|
| area of XS \(= \frac{1}{2} \times (8x + 10x) \times x = 9x^2\) | M1 |
| volume \(= 9x^2y = 900\) | M1 |
| \(\therefore y = \frac{100}{x^2}\) | A1 |
| Answer | Marks |
|---|---|
| width of sloping sides \(= \sqrt{2}x\) | B1 |
| \(A = 8xy + 2(9x^2) + 2(\sqrt{2}xy)\) | M1 |
| \(A = 18x^2 + 2xy(4 + \sqrt{2})\) | |
| \(A = 18x^2 + 2x(4 + \sqrt{2}) \times \frac{100}{x^2}\) | M1 |
| \(A = 18x^2 + \frac{200(4 + \sqrt{2})}{x}\) | A1 |
| Answer | Marks |
|---|---|
| \(\frac{dA}{dx} = 36x - 200(4 + \sqrt{2})x^{-2}\) | M1 A1 |
| for SP, \(36x - 200(4 + \sqrt{2})x^{-2} = 0\) | M1 |
| \(x^3 = \frac{200(4 + \sqrt{2})}{36}\) | |
| \(x = \sqrt[3]{\frac{50(4 + \sqrt{2})}{9}} = 3.11\) | A1 |
| Answer | Marks |
|---|---|
| \(A = 522\) (3sf) | B1 |
| \(\frac{d^2A}{dx^2} = 36 + 400(4 + \sqrt{2})x^{-3}\) | M1 |
| when \(x = 3.11\), \(\frac{d^2A}{dx^2} = 108\), \(\frac{d^2A}{dx^2} > 0 \therefore\) minimum | A1 |
**Part (a)**
area of XS $= \frac{1}{2} \times (8x + 10x) \times x = 9x^2$ | M1 |
volume $= 9x^2y = 900$ | M1 |
$\therefore y = \frac{100}{x^2}$ | A1 |
**Part (b)**
width of sloping sides $= \sqrt{2}x$ | B1 |
$A = 8xy + 2(9x^2) + 2(\sqrt{2}xy)$ | M1 |
$A = 18x^2 + 2xy(4 + \sqrt{2})$ |
$A = 18x^2 + 2x(4 + \sqrt{2}) \times \frac{100}{x^2}$ | M1 |
$A = 18x^2 + \frac{200(4 + \sqrt{2})}{x}$ | A1 |
**Part (c)**
$\frac{dA}{dx} = 36x - 200(4 + \sqrt{2})x^{-2}$ | M1 A1 |
for SP, $36x - 200(4 + \sqrt{2})x^{-2} = 0$ | M1 |
$x^3 = \frac{200(4 + \sqrt{2})}{36}$ |
$x = \sqrt[3]{\frac{50(4 + \sqrt{2})}{9}} = 3.11$ | A1 |
**Part (d)**
$A = 522$ (3sf) | B1 |
$\frac{d^2A}{dx^2} = 36 + 400(4 + \sqrt{2})x^{-3}$ | M1 |
when $x = 3.11$, $\frac{d^2A}{dx^2} = 108$, $\frac{d^2A}{dx^2} > 0 \therefore$ minimum | A1 |
---9.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{9215e382-406c-41a3-8907-f465b134dd87-4_499_1137_954_319}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows a tray made from sheet metal.\\
The horizontal base is a rectangle measuring $8 x \mathrm {~cm}$ by $y \mathrm {~cm}$ and the two vertical sides are trapezia of height $x \mathrm {~cm}$ with parallel edges of length $8 x \mathrm {~cm}$ and $10 x \mathrm {~cm}$. The remaining two sides are rectangles inclined at $45 ^ { \circ }$ to the horizontal.
Given that the capacity of the tray is $900 \mathrm {~cm} ^ { 3 }$,
\begin{enumerate}[label=(\alph*)]
\item find an expression for $y$ in terms of $x$,
\item show that the area of metal used to make the tray, $A \mathrm {~cm} ^ { 2 }$, is given by
$$A = 18 x ^ { 2 } + \frac { 200 ( 4 + \sqrt { 2 } ) } { x } ,$$
\item find to 3 significant figures, the value of $x$ for which $A$ is stationary,
\item find the minimum value of $A$ and show that it is a minimum.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 Q9 [14]}}