| Exam Board | AQA |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Year | 2021 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Stationary points and optimisation |
| Type | Open box from cut-corner sheet |
| Difficulty | Moderate -0.3 This is a classic optimization problem with a standard setup (box from cut corners). Part (a) requires straightforward volume calculation V=length×width×height with given dimensions. Part (b) involves routine differentiation, solving dC/dx=0, and using second derivative test. While multi-step, it follows a well-practiced template with no novel insight required, making it slightly easier than average. |
| Spec | 1.07n 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 |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Side of box is \(30 - 2x\) | M1 (3.1b) | |
| \(C = x(30-2x)^2 = x(900 - 120x + 4x^2)\) leading to \(C = 900x - 120x^2 + 4x^3\) | R1 (2.1) | Must identify three dimensions then expand correctly to obtain answer AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dC}{dx} = 900 - 240x + 12x^2\) (at least one term correct) | M1 (1.1a) | |
| Fully correct derivative | A1 (1.1b) | |
| For maximum \(\frac{dC}{dx} = 0\) | E1 (2.4) | Must explain \(\frac{dC}{dx}\) must be 0 for turning point |
| Equate derivative to 0, solve for \(x < 15\): \(x = 5\) or \(x = 15\), however \(x < 15\) so \(x = 5\) | M1 (1.1a) | |
| \(x = 5\) | A1 (1.1b) | |
| \(\frac{d^2C}{dx^2} = -240 + 24x\), negative when \(x = 5\) | R1 (2.1) | Justifies \(x = 5\) is the maximum |
| Maximum capacity \(C = 2000 \text{ cm}^3\) | A1 (3.2a) | CAO; condone incorrect or missing units |
## Question 10(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Side of box is $30 - 2x$ | M1 (3.1b) | |
| $C = x(30-2x)^2 = x(900 - 120x + 4x^2)$ leading to $C = 900x - 120x^2 + 4x^3$ | R1 (2.1) | Must identify three dimensions then expand correctly to obtain answer AG |
## Question 10(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dC}{dx} = 900 - 240x + 12x^2$ (at least one term correct) | M1 (1.1a) | |
| Fully correct derivative | A1 (1.1b) | |
| For maximum $\frac{dC}{dx} = 0$ | E1 (2.4) | Must explain $\frac{dC}{dx}$ must be 0 for turning point |
| Equate derivative to 0, solve for $x < 15$: $x = 5$ or $x = 15$, however $x < 15$ so $x = 5$ | M1 (1.1a) | |
| $x = 5$ | A1 (1.1b) | |
| $\frac{d^2C}{dx^2} = -240 + 24x$, negative when $x = 5$ | R1 (2.1) | Justifies $x = 5$ is the maximum |
| Maximum capacity $C = 2000 \text{ cm}^3$ | A1 (3.2a) | CAO; condone incorrect or missing units |10 A square sheet of metal has edges 30 cm long.
Four squares each with edge $x \mathrm {~cm}$, where $x < 15$, are removed from the corners of the sheet.
The four rectangular sections are bent upwards to form an open-topped box, as shown in the diagrams.\\
\includegraphics[max width=\textwidth, alt={}, center]{f87d1b36-26db-4a0b-b9ec-d7d82a396aba-12_392_460_630_347}\\
\includegraphics[max width=\textwidth, alt={}, center]{f87d1b36-26db-4a0b-b9ec-d7d82a396aba-12_387_437_635_872}\\
\includegraphics[max width=\textwidth, alt={}, center]{f87d1b36-26db-4a0b-b9ec-d7d82a396aba-12_282_380_703_1318}
10
\begin{enumerate}[label=(\alph*)]
\item Show that the capacity, $C \mathrm {~cm} ^ { 3 }$, of the box is given by
$$C = 900 x - 120 x ^ { 2 } + 4 x ^ { 3 }$$
10
\item Find the maximum capacity of the box.
Fully justify your answer.
\end{enumerate}
\hfill \mbox{\textit{AQA AS Paper 2 2021 Q10 [9]}}