| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Session | Specimen |
| Marks | 9 |
| 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-to-surface-area substitution (part a), differentiation and solving dS/dr=0 (part b), and a contextual comment (part c). The algebra is straightforward with no novel insight required, making it slightly easier than average for A-level. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.07t Construct differential equations: in context |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Sets \(500 = \pi r^2 h\) | B1 | Uses correct volume formula with \(V=500\) |
| Substitute \(h = \frac{500}{\pi r^2}\) into \(S = 2\pi r^2 + 2\pi r h = 2\pi r^2 + 2\pi r \times \frac{500}{\pi r^2}\) | M1 | Substitutes \(h = \frac{500}{\pi r^2}\) or \(rh = \frac{500}{\pi r}\) into \(S = 2\pi r^2 + 2\pi rh\) to get \(S\) as a function of \(r\) |
| Simplifies to reach given answer \(S = 2\pi r^2 + \frac{1000}{r}\) | A1* | Given answer — must be shown |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Differentiates \(S\) with both indices correct in \(\frac{dS}{dr}\) | M1 | Differentiates to form \(Ar \pm Br^{-2}\) |
| \(\frac{dS}{dr} = 4\pi r - \frac{1000}{r^2}\) | A1 | Or exact equivalent |
| Sets \(\frac{dS}{dr} = 0\) and proceeds to \(r^3 = k\), \(k\) is a constant | M1 | |
| Radius \(= 4.30\) cm | A1 | \(R =\) awrt \(4.30\) cm |
| Substitutes \(r = 4.30\) into \(h = \frac{500}{\pi r^2} \Rightarrow\) Height \(= 8.60\) cm | A1 | \(H =\) awrt \(8.60\) cm |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| States a valid reason e.g. radius too big for hands; \(r=4.3\) cm and \(h=8.6\) cm makes can square in profile; radius too big to drink from; different dimensions to other cans; difficult to stack | B1 | Any valid reason |
## Question 14:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Sets $500 = \pi r^2 h$ | B1 | Uses correct volume formula with $V=500$ |
| Substitute $h = \frac{500}{\pi r^2}$ into $S = 2\pi r^2 + 2\pi r h = 2\pi r^2 + 2\pi r \times \frac{500}{\pi r^2}$ | M1 | Substitutes $h = \frac{500}{\pi r^2}$ or $rh = \frac{500}{\pi r}$ into $S = 2\pi r^2 + 2\pi rh$ to get $S$ as a function of $r$ |
| Simplifies to reach given answer $S = 2\pi r^2 + \frac{1000}{r}$ | A1* | Given answer — must be shown |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Differentiates $S$ with both indices correct in $\frac{dS}{dr}$ | M1 | Differentiates to form $Ar \pm Br^{-2}$ |
| $\frac{dS}{dr} = 4\pi r - \frac{1000}{r^2}$ | A1 | Or exact equivalent |
| Sets $\frac{dS}{dr} = 0$ and proceeds to $r^3 = k$, $k$ is a constant | M1 | |
| Radius $= 4.30$ cm | A1 | $R =$ awrt $4.30$ cm |
| Substitutes $r = 4.30$ into $h = \frac{500}{\pi r^2} \Rightarrow$ Height $= 8.60$ cm | A1 | $H =$ awrt $8.60$ cm |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| States a valid reason e.g. radius too big for hands; $r=4.3$ cm and $h=8.6$ cm makes can square in profile; radius too big to drink from; different dimensions to other cans; difficult to stack | B1 | Any valid reason |
---\begin{enumerate}
\item A company decides to manufacture a soft drinks can with a capacity of 500 ml .
\end{enumerate}
The company models the can in the shape of a right circular cylinder with radius $r \mathrm {~cm}$ and height $h \mathrm {~cm}$.
In the model they assume that the can is made from a metal of negligible thickness.\\
(a) Prove that the total surface area, $S \mathrm {~cm} ^ { 2 }$, of the can is given by
$$S = 2 \pi r ^ { 2 } + \frac { 1000 } { r }$$
Given that $r$ can vary,\\
(b) find the dimensions of a can that has minimum surface area.\\
(c) With reference to the shape of the can, suggest a reason why the company may choose not to manufacture a can with minimum surface area.
\hfill \mbox{\textit{Edexcel Paper 2 Q14 [9]}}