| Exam Board | WJEC |
|---|---|
| Module | Unit 1 (Unit 1) |
| Session | Specimen |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Applied differentiation |
| Type | Optimise 3D shape dimensions |
| Difficulty | Standard +0.3 This is a standard optimization problem requiring volume constraint substitution and basic calculus (differentiation, finding stationary points, second derivative test). The algebra is straightforward and the problem follows a familiar textbook pattern with clear guidance, making it slightly easier than average. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives |
| Answer | Marks | Guidance |
|---|---|---|
| (a) Height of box \(= \frac{9000}{2x^2}\) | B1 | AO3 (o.e.) |
| \(S = 2 \times (2x \times x + \frac{9000}{2x^2} \times x + \frac{9000}{2x^2} \times 2x)\) | M1 | AO3 (f.t. candidate's derived expression for height of box in terms of x) |
| \(S = 4x^2 + \frac{27000}{x}\) | A1 | AO3 (convincing) |
| (b) \(\frac{dS}{dx} = 8x – \frac{27000}{x^2}\) | B1 | AO1 |
| Putting derived \(\frac{dS}{dx} = 0\) | M1 | AO1 |
| \(x = 15\) | A1 | AO1 (f.t. candidate's \(\frac{dS}{dx}\)) |
| Stationary value of \(S\) at \(x = 15\) is 2700 | A1 | AO1 (c.a.o.) |
| A correct method for finding nature of the stationary point yielding a minimum value | B1 | AO1 |
**(a)** Height of box $= \frac{9000}{2x^2}$ | B1 | AO3 (o.e.)
$S = 2 \times (2x \times x + \frac{9000}{2x^2} \times x + \frac{9000}{2x^2} \times 2x)$ | M1 | AO3 (f.t. candidate's derived expression for height of box in terms of x)
$S = 4x^2 + \frac{27000}{x}$ | A1 | AO3 (convincing)
**(b)** $\frac{dS}{dx} = 8x – \frac{27000}{x^2}$ | B1 | AO1
Putting derived $\frac{dS}{dx} = 0$ | M1 | AO1
$x = 15$ | A1 | AO1 (f.t. candidate's $\frac{dS}{dx}$)
Stationary value of $S$ at $x = 15$ is 2700 | A1 | AO1 (c.a.o.)
A correct method for finding nature of the stationary point yielding a minimum value | B1 | AO1
**Total: [8]**
---The diagram below shows a closed box in the form of a cuboid, which is such that the length of its base is twice the width of its base. The volume of the box is 9000 cm³. The total surface area of the box is denoted by $S$ cm².
\includegraphics{figure_14}
\begin{enumerate}[label=(\alph*)]
\item Show that $S = 4x^2 + \frac{27000}{x}$, where $x$ cm denotes the width of the base. [3]
\item Find the minimum value of $S$, showing that the value you have found is a minimum value. [5]
\end{enumerate}
\hfill \mbox{\textit{WJEC Unit 1 Q14 [8]}}