| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2015 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Stationary points and optimisation |
| Type | Show formula then optimise: cost/non-geometric objective |
| Difficulty | Standard +0.3 This is a standard C2 optimization problem with a given constraint. Part (a) requires substituting the volume constraint into the cost formula (routine algebra), part (b) involves differentiating and solving dC/dr = 0 (standard technique), and part (c) asks for second derivative test verification. While multi-step, each component is a textbook application of differentiation with no novel insight required, making it slightly easier than average. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.07p Points of inflection: using second derivative1.07t Construct differential equations: in context |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| States \(3 \times 2\pi r^2\) or \(2 \times 2\pi r h\) | B1 | At least one correct product seen |
| \(h = \frac{75\pi}{\pi r^2}\) or \(h = \frac{75}{r^2}\) | B1ft | Correct expression for \(h\) in terms of \(r\) |
| \(C = 6\pi r^2 + 4\pi r\left(\frac{75}{r^2}\right)\) | M1 | Substitutes expression for \(h\) into area/cost of form \(Ar^2 + Brh\) |
| \(C = 6\pi r^2 + \frac{300\pi}{r}\) | A1* | Given answer; no errors seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{dC}{dr} = 12\pi r - \frac{300\pi}{r^2}\) or \(12\pi r - 300\pi r^{-2}\) | M1 A1ft | Attempt to differentiate; correct derivative |
| \(12\pi r - \frac{300\pi}{r^2} = 0\) so \(r^k =\) value where \(k = \pm 2, \pm 3, \pm 4\) | dM1 | Sets \(\frac{dC}{dr} = 0\) |
| Uses cube root to obtain \(r = \left(\frac{300}{12}\right)^{\frac{1}{3}} = 2.92\) | ddM1 | Uses cube root; allow \(r=3\) as evidence |
| \(C =\) awrt 483 or 484 | A1cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{d^2C}{dr^2} = 12\pi + \frac{600\pi}{r^3} > 0\) so minimum | B1ft | Correct second derivative and correct conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| As before | B1 | |
| Uses volume to give \(h = \frac{75}{\pi r^2}\) | B1 | |
| \(C = 6\pi r^2 + 4\pi r\left(\frac{75}{\pi r^2}\right)\) | M1 | |
| Printed answer not obtained without error | A0 | Maximum award: B1 B1 M1 A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{dC}{dr} = 12\pi r - \frac{300}{r^2}\) or \(12\pi r - 300r^{-2}\) | M1 A1 | |
| \(12\pi r - \frac{300}{r^2} = 0\) so \(r^k\) = value, where \(k = 2, 3\) or \(4\) | dM1 | |
| Use cube root to obtain \(r = \left(\frac{300}{12\pi}\right)^{\frac{1}{3}}\) \(= 1.996\), allow \(r = 2\), then find \(C\) | ddM1 | Must use \(C = 6\pi r^2 + \frac{300}{r}\) |
| Cannot obtain \(C = 483\) or \(484\) | A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{d^2C}{dr^2} = 12\pi + \frac{600}{r^3} > 0\) so minimum | B1 | OR checks gradient either side of \(1.966\) showing negative to positive; OR checks value of \(C\) either side of \(1.966\) showing \(C > 225.4\), deducing minimum via graph shape |
# Question 9:
## Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| States $3 \times 2\pi r^2$ or $2 \times 2\pi r h$ | B1 | At least one correct product seen |
| $h = \frac{75\pi}{\pi r^2}$ or $h = \frac{75}{r^2}$ | B1ft | Correct expression for $h$ in terms of $r$ |
| $C = 6\pi r^2 + 4\pi r\left(\frac{75}{r^2}\right)$ | M1 | Substitutes expression for $h$ into area/cost of form $Ar^2 + Brh$ |
| $C = 6\pi r^2 + \frac{300\pi}{r}$ | A1* | Given answer; no errors seen |
## Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{dC}{dr} = 12\pi r - \frac{300\pi}{r^2}$ or $12\pi r - 300\pi r^{-2}$ | M1 A1ft | Attempt to differentiate; correct derivative |
| $12\pi r - \frac{300\pi}{r^2} = 0$ so $r^k =$ value where $k = \pm 2, \pm 3, \pm 4$ | dM1 | Sets $\frac{dC}{dr} = 0$ |
| Uses cube root to obtain $r = \left(\frac{300}{12}\right)^{\frac{1}{3}} = 2.92$ | ddM1 | Uses cube root; allow $r=3$ as evidence |
| $C =$ awrt 483 or 484 | A1cao | |
## Part (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{d^2C}{dr^2} = 12\pi + \frac{600\pi}{r^3} > 0$ so minimum | B1ft | Correct second derivative and correct conclusion |
## Misread Version (C = 6πr² + 300/r)
**Part (a) - Misread version:**
| Working/Answer | Mark | Guidance |
|---|---|---|
| As before | B1 | |
| Uses volume to give $h = \frac{75}{\pi r^2}$ | B1 | |
| $C = 6\pi r^2 + 4\pi r\left(\frac{75}{\pi r^2}\right)$ | M1 | |
| Printed answer not obtained without error | A0 | Maximum award: B1 B1 M1 A0 |
---
**Part (b) - Misread version:**
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{dC}{dr} = 12\pi r - \frac{300}{r^2}$ or $12\pi r - 300r^{-2}$ | M1 A1 | |
| $12\pi r - \frac{300}{r^2} = 0$ so $r^k$ = value, where $k = 2, 3$ or $4$ | dM1 | |
| Use cube root to obtain $r = \left(\frac{300}{12\pi}\right)^{\frac{1}{3}}$ $= 1.996$, allow $r = 2$, then find $C$ | ddM1 | Must use $C = 6\pi r^2 + \frac{300}{r}$ |
| Cannot obtain $C = 483$ or $484$ | A0 | |
---
**Part (c) - Misread version:**
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{d^2C}{dr^2} = 12\pi + \frac{600}{r^3} > 0$ **so minimum** | B1 | OR checks gradient either side of $1.966$ showing negative to positive; OR checks value of $C$ either side of $1.966$ showing $C > 225.4$, deducing minimum via graph shape |9. A solid glass cylinder, which is used in an expensive laser amplifier, has a volume of $75 \pi \mathrm {~cm} ^ { 3 }$.\\
The cost of polishing the surface area of this glass cylinder is $\pounds 2$ per $\mathrm { cm } ^ { 2 }$ for the curved surface area and $\pounds 3$ per $\mathrm { cm } ^ { 2 }$ for the circular top and base areas.
Given that the radius of the cylinder is $r \mathrm {~cm}$,
\begin{enumerate}[label=(\alph*)]
\item show that the cost of the polishing, $\pounds C$, is given by
$$C = 6 \pi r ^ { 2 } + \frac { 300 \pi } { r }$$
\item Use calculus to find the minimum cost of the polishing, giving your answer to the nearest pound.
\item Justify that the answer that you have obtained in part (b) is a minimum.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2015 Q9 [10]}}