| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2016 |
| Session | March |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chain Rule |
| Type | Optimization with constraint |
| Difficulty | Standard +0.3 This is a standard optimization problem requiring volume constraint substitution, differentiation using power rule, and verification of minimum via second derivative. While multi-step, it follows a routine textbook template with clear guidance ('show that') and involves only basic calculus techniques—slightly easier than average A-level questions. |
| 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 derivatives1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(A = 2\pi r^2 + 2\pi rh\) | B1 | |
| \(\pi r^2 h = 1000 \rightarrow h = \dfrac{1000}{\pi r^2}\) | M1 | |
| Sub for \(h\) into \(A \rightarrow A = 2\pi r^2 + \dfrac{2000}{r}\) AG | A1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\dfrac{dA}{dr} = 0 \Rightarrow 4\pi r - \dfrac{2000}{r^2} = 0\) | M1A1 | Attempt differentiation & set \(= 0\) |
| \(r = 5.4\) | DM1 A1 | Reasonable attempt to solve to \(r^3 =\) |
| \(\dfrac{d^2A}{dr^2} = 4\pi + \dfrac{4000}{r^3}\) | ||
| \(> 0\) hence MIN hence MOST EFFICIENT AG | B1 [5] | Or convincing alternative method |
## Question 6:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $A = 2\pi r^2 + 2\pi rh$ | B1 | |
| $\pi r^2 h = 1000 \rightarrow h = \dfrac{1000}{\pi r^2}$ | M1 | |
| Sub for $h$ into $A \rightarrow A = 2\pi r^2 + \dfrac{2000}{r}$ **AG** | A1 [3] | |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{dA}{dr} = 0 \Rightarrow 4\pi r - \dfrac{2000}{r^2} = 0$ | M1A1 | Attempt differentiation & set $= 0$ |
| $r = 5.4$ | DM1 A1 | Reasonable attempt to solve to $r^3 =$ |
| $\dfrac{d^2A}{dr^2} = 4\pi + \dfrac{4000}{r^3}$ | | |
| $> 0$ hence MIN hence MOST EFFICIENT **AG** | B1 [5] | Or convincing alternative method |
---6 A vacuum flask (for keeping drinks hot) is modelled as a closed cylinder in which the internal radius is $r \mathrm {~cm}$ and the internal height is $h \mathrm {~cm}$. The volume of the flask is $1000 \mathrm {~cm} ^ { 3 }$. A flask is most efficient when the total internal surface area, $A \mathrm {~cm} ^ { 2 }$, is a minimum.\\
(i) Show that $A = 2 \pi r ^ { 2 } + \frac { 2000 } { r }$.\\
(ii) Given that $r$ can vary, find the value of $r$, correct to 1 decimal place, for which $A$ has a stationary value and verify that the flask is most efficient when $r$ takes this value.
\hfill \mbox{\textit{CAIE P1 2016 Q6 [8]}}