| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Stationary points and optimisation |
| Type | Show formula then optimise: cylinder/prism (single variable) |
| Difficulty | Standard +0.3 This is a standard optimization problem with guided steps. Part (i) requires setting up a constraint equation (surface area) and substituting to eliminate h—a routine algebraic manipulation. Parts (ii)-(iv) involve differentiating, finding stationary points, and using the second derivative test, all standard C1 techniques. The problem is slightly easier than average because it's heavily scaffolded with the formula given in part (i), requiring no independent modeling. |
| Spec | 1.02z Models in context: use functions in modelling1.07n 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 | Notes |
| \(A = \pi r^2 + 2\pi r h = 192\pi\) | M1 | |
| \(\therefore h = \frac{192 - r^2}{2r} = \frac{96}{r} - \frac{r}{2}\) | M1 A1 | |
| \(V = \pi r^2 h = \pi r^2\left(\frac{96}{r} - \frac{r}{2}\right) = 96\pi r - \frac{1}{2}\pi r^3\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(\frac{\mathrm{d}V}{\mathrm{d}r} = 96\pi - \frac{3}{2}\pi r^2\) | M1 A1 | |
| For SP, \(96\pi - \frac{3}{2}\pi r^2 = 0\) | M1 | |
| \(r^2 = 64\) | ||
| \(r = 8\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(V = (96\pi \times 8) - \left(\frac{1}{2}\pi \times 8^3\right) = 512\pi\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(\frac{\mathrm{d}^2V}{\mathrm{d}r^2} = -3\pi r\) | M1 | |
| When \(r = 8\), \(\frac{\mathrm{d}^2V}{\mathrm{d}r^2} = -24\pi\), \(\frac{\mathrm{d}^2V}{\mathrm{d}r^2} < 0\) \(\therefore\) maximum | A1 | (13) |
# Question 10:
## Part (i)
| Answer/Working | Marks | Notes |
|---|---|---|
| $A = \pi r^2 + 2\pi r h = 192\pi$ | M1 | |
| $\therefore h = \frac{192 - r^2}{2r} = \frac{96}{r} - \frac{r}{2}$ | M1 A1 | |
| $V = \pi r^2 h = \pi r^2\left(\frac{96}{r} - \frac{r}{2}\right) = 96\pi r - \frac{1}{2}\pi r^3$ | M1 A1 | |
## Part (ii)
| Answer/Working | Marks | Notes |
|---|---|---|
| $\frac{\mathrm{d}V}{\mathrm{d}r} = 96\pi - \frac{3}{2}\pi r^2$ | M1 A1 | |
| For SP, $96\pi - \frac{3}{2}\pi r^2 = 0$ | M1 | |
| $r^2 = 64$ | | |
| $r = 8$ | A1 | |
## Part (iii)
| Answer/Working | Marks | Notes |
|---|---|---|
| $V = (96\pi \times 8) - \left(\frac{1}{2}\pi \times 8^3\right) = 512\pi$ | M1 A1 | |
## Part (iv)
| Answer/Working | Marks | Notes |
|---|---|---|
| $\frac{\mathrm{d}^2V}{\mathrm{d}r^2} = -3\pi r$ | M1 | |
| When $r = 8$, $\frac{\mathrm{d}^2V}{\mathrm{d}r^2} = -24\pi$, $\frac{\mathrm{d}^2V}{\mathrm{d}r^2} < 0$ $\therefore$ maximum | A1 | **(13)** |
**Total: (72)**10.\\
\includegraphics[max width=\textwidth, alt={}, center]{6ef55dbd-f18d-4264-b80c-d181473ca7b3-3_531_786_246_523}
The diagram shows an open-topped cylindrical container made from cardboard. The cylinder is of height $h \mathrm {~cm}$ and base radius $r \mathrm {~cm}$.
Given that the area of card used to make the container is $192 \pi \mathrm {~cm} ^ { 2 }$,\\
(i) show that the capacity of the container, $\mathrm { V } \mathrm { cm } ^ { 3 }$, is given by
$$V = 96 \pi r - \frac { 1 } { 2 } \pi r ^ { 3 } .$$
(ii) Find the value of $r$ for which $V$ is stationary.\\
(iii) Find the corresponding value of $V$ in terms of $\pi$.\\
(iv) Determine whether this is a maximum or a minimum value of $V$.
\hfill \mbox{\textit{OCR C1 Q10 [13]}}