| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2006 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Applied differentiation |
| Type | Optimise 3D shape dimensions |
| Difficulty | Moderate -0.3 This is a standard optimization problem with clear scaffolding: deriving a formula (given the answer), differentiating a simple expression, then finding and justifying a minimum. All steps are routine C1 techniques with no novel insight required. Slightly easier than average due to the scaffolding in part (i), though the justification in part (iii) adds minor challenge. |
| Spec | 1.02z Models in context: use functions in modelling1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives |
| Answer | Marks | Guidance |
|---|---|---|
| Height of box = \(\frac{8}{x^2}\) | *B1 | Area of 1 vertical face = \(\frac{8}{x^2} \times x = \frac{8}{x}\) |
| 4 vertical faces = \(4 \times \frac{8}{x} = \frac{32}{x}\) | *B1 | |
| Total surface area = \(x^2 + x^2 + \frac{32}{x}\) | B1 dep on both ** | Correct final expression; 3 marks |
| \(A = 2x^2 + \frac{32}{x}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dA}{dx} = 4x - \frac{32}{x^2}\) | B1 | \(4x\) |
| \(\frac{dA}{dx} = 4x - \frac{32}{x^2}\) | B1 | \(kx^2\); 3 marks |
| B1 | \(-32x^2\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(4x - \frac{32}{x^2} = 0\) | M1 | \(\frac{dA}{dx} = 0\) soi |
| \(4x^3 = 32\) | M1 | |
| \(x = 2\) | A1 | |
| Check for minimum; Correctly justified | M1 | 4 marks |
| A1 | ||
| SR If \(x = 2\) stated www but with no evidence of differentiated expression(s) having been used in part (iii) | B1 |
## (i)
Height of box = $\frac{8}{x^2}$ | *B1 | Area of 1 vertical face = $\frac{8}{x^2} \times x = \frac{8}{x}$
4 vertical faces = $4 \times \frac{8}{x} = \frac{32}{x}$ | *B1 |
Total surface area = $x^2 + x^2 + \frac{32}{x}$ | B1 dep on both ** | Correct final expression; 3 marks
$A = 2x^2 + \frac{32}{x}$ | |
## (ii)
$\frac{dA}{dx} = 4x - \frac{32}{x^2}$ | B1 | $4x$
$\frac{dA}{dx} = 4x - \frac{32}{x^2}$ | B1 | $kx^2$; 3 marks
| B1 | $-32x^2$
## (iii)
$4x - \frac{32}{x^2} = 0$ | M1 | $\frac{dA}{dx} = 0$ soi
$4x^3 = 32$ | M1 |
$x = 2$ | A1 |
Check for minimum; Correctly justified | M1 | 4 marks
| A1 |
SR If $x = 2$ stated www but with no evidence of differentiated expression(s) having been used in part (iii) | B1 |
---A cuboid has a volume of $8 \text{m}^3$. The base of the cuboid is square with sides of length $x$ metres. The surface area of the cuboid is $A \text{m}^2$.
\begin{enumerate}[label=(\roman*)]
\item Show that $A = 2x^2 + \frac{32}{x}$. [3]
\item Find $\frac{dA}{dx}$. [3]
\item Find the value of $x$ which gives the smallest surface area of the cuboid, justifying your answer. [4]
\end{enumerate}
\hfill \mbox{\textit{OCR C1 2006 Q8 [10]}}