| Exam Board | OCR MEI |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2011 |
| Session | June |
| Marks | 11 |
| 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: part (i) guides students through algebraic manipulation to eliminate h, part (ii) is routine differentiation, and part (iii) applies the standard maximum-finding procedure. While it requires multiple steps, each step follows a well-practiced technique with no novel insight needed, making it slightly easier than average. |
| 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 |
|---|---|---|
| (i) \(200 - 2\pi r^2 = 2\pi rh\) | M1 | \(100 = \pi r^2 + \pi rh\) |
| \(h = \frac{200 - 2\pi r^2}{2\pi r}\) o.e. | M1 | \(100r = \pi r^3 + \pi r^2 h\) |
| substitution of correct \(h\) into \(V = \pi r^2 h\) | M1 | \(100r = \pi r^3 + V\) |
| \(V = 100r - \pi r^3\) convincingly obtained | A1 | \(V = 100r - \pi r^3\) or M1 for \(h = \frac{V}{\pi r^2}\); M1 for \(200 = 2\pi r^2 + 2\pi r \times \frac{V}{\pi r^2}\); M1 for \(200 = 2\pi r^2 + 2\frac{V}{r}\); A1 for \(V = 100r - \pi r^3\) convincingly obtained |
| (ii) \(\frac{dV}{dr} = 100 - 3\pi r^2\) | B2 | B1 for each term |
| \(\frac{d^2V}{dr^2} = -6\pi r\) | B1 | \(-18.8(…) r\) or better if decimalised |
| (iii) their \(\frac{dV}{dr} = 0\) s.o.i. | M1 | must contain \(r\) as the only variable |
| \(r = 3.26\) c.a.o. | A2 | A1 for \(r = (\pm)\sqrt{\frac{100}{3\pi}}\); may be implied by 3.25… |
| \(V = 217\) c.a.o. | A1 | deduct 1 mark only in this part if answers not given to 3 sf; there must be evidence of use of calculus |
**(i)** $200 - 2\pi r^2 = 2\pi rh$ | M1 | $100 = \pi r^2 + \pi rh$
$h = \frac{200 - 2\pi r^2}{2\pi r}$ o.e. | M1 | $100r = \pi r^3 + \pi r^2 h$
substitution of correct $h$ into $V = \pi r^2 h$ | M1 | $100r = \pi r^3 + V$
$V = 100r - \pi r^3$ convincingly obtained | A1 | $V = 100r - \pi r^3$ or M1 for $h = \frac{V}{\pi r^2}$; M1 for $200 = 2\pi r^2 + 2\pi r \times \frac{V}{\pi r^2}$; M1 for $200 = 2\pi r^2 + 2\frac{V}{r}$; A1 for $V = 100r - \pi r^3$ convincingly obtained | sc3 for complete argument working backwards: $V = 100r - \pi r^3$; $\pi r^2 h = 100r - \pi r^3$; $\pi r^2 h = 100r - \pi r^3$; $\pi rh = 100 - \pi r^2$; $100 = \pi rh + \pi r^2$; $200 = A = 2\pi rh + 2\pi r^2$; sc0 if argument is incomplete
**(ii)** $\frac{dV}{dr} = 100 - 3\pi r^2$ | B2 | B1 for each term | allow $9.42(….) r^2$ or better if decimalised
$\frac{d^2V}{dr^2} = -6\pi r$ | B1 | $-18.8(…) r$ or better if decimalised
**(iii)** their $\frac{dV}{dr} = 0$ s.o.i. | M1 | must contain $r$ as the only variable
$r = 3.26$ c.a.o. | A2 | A1 for $r = (\pm)\sqrt{\frac{100}{3\pi}}$; may be implied by 3.25…
$V = 217$ c.a.o. | A1 | deduct 1 mark only in this part if answers not given to 3 sf; there must be evidence of use of calculus11 (i) The standard formulae for the volume $V$ and total surface area $A$ of a solid cylinder of radius $r$ and height $h$ are
$$V = \pi r ^ { 2 } h \quad \text { and } \quad A = 2 \pi r ^ { 2 } + 2 \pi r h .$$
Use these to show that, for a cylinder with $A = 200$,
$$V = 100 r - \pi r ^ { 3 }$$
(ii) Find $\frac { \mathrm { d } V } { \mathrm {~d} r }$ and $\frac { \mathrm { d } ^ { 2 } V } { \mathrm {~d} r ^ { 2 } }$.\\
(iii) Use calculus to find the value of $r$ that gives a maximum value for $V$ and hence find this maximum value, giving your answers correct to 3 significant figures.
\hfill \mbox{\textit{OCR MEI C2 2011 Q11 [11]}}