| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2022 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Stationary points and optimisation |
| Type | Optimise cost or profit model |
| Difficulty | Standard +0.3 This is a standard AS-level optimization problem with clear scaffolding through parts (a)-(d). Students must form a cost function using given constraints, differentiate to find critical points, and verify the minimum using the second derivative test. While it requires multiple techniques (constraint handling, differentiation, second derivative test), each step is routine and the question structure guides students through the solution methodically. Slightly easier than average due to the scaffolding and standard nature of the problem. |
| 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 |
|---|---|---|
| Working | Mark | Guidance |
| \(V=\pi r^2 h = 355 \Rightarrow h=\frac{355}{\pi r^2}\) (or \(rh=\frac{355}{\pi r}\) or \(\pi rh=\frac{355}{r}\)) | B1 | 1.1b — Correct expression for \(h\) or \(rh\) or \(\pi rh\) in terms of \(r\); may be implied by later substitution |
| \(C=0.04(\pi r^2+2\pi rh)+0.09(\pi r^2)\) | M1 | 3.4 — Scored for sum of three terms of form \(0.04...r^2\), \(0.09...r^2\) and \(0.04\times...rh\) |
| \(C=0.13\pi r^2+0.08\pi rh=0.13\pi r^2+0.08\pi r\left(\frac{355}{\pi r^2}\right)\) | dM1 | 2.1 — Substitutes \(h\) into \(C\); dependent on correct form of \(h\) |
| \(C=0.13\pi r^2+\frac{28.4}{r}\) | A1* | 1.1b — Fully correct, no errors. Must see: separate volume equation, separate costs before combining, substitution before \(\frac{28.4}{r}\) term |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\frac{dC}{dr}=0.26\pi r - \frac{28.4}{r^2}\) | M1, A1 | 3.4, 1.1b — M1: differentiates to get at least \(r^{-1}\to r^{-2}\); A1: correct derivative |
| \(\frac{dC}{dr}=0 \Rightarrow r^3=\frac{28.4}{0.26\pi} \Rightarrow r=...\) | M1 | 1.1b — Sets \(\frac{dC}{dr}=0\), solves for \(r\); do not withhold if \(r\) is negative |
| \(r=\sqrt[3]{\frac{1420}{13\pi}}=3.26...\) | A1 | 1.1b — Correct value; allow exact or awrt 3.26 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\frac{d^2C}{dr^2}=0.26\pi+\frac{56.8}{r^3}=0.26\pi+\frac{56.8}{\text{"3.26"}^3}\) | M1 | 1.1b — Finds \(\frac{d^2C}{dr^2}\) at their positive \(r\); valid if second derivative is of form \(A+\frac{B}{r^3}\) where \(A,B\neq 0\) |
| \(\frac{d^2C}{dr^2}=(2.45...)>0\), hence minimum (cost) | A1 | 2.4 — Requires correct \(\frac{d^2C}{dr^2}\) AND deduction that \(\frac{d^2C}{dr^2}>0\) with minimal explanation |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(C=0.13\pi(\text{"3.26"})^2+\frac{28.4}{\text{"3.26"}}\) | M1 | 3.4 — Uses model with positive \(r\) from (b); may be seen in (b) |
| \((C=)13\) | A1 | 1.1b — Ignore units |
# Question 12:
**Part (a):**
| Working | Mark | Guidance |
|---------|------|----------|
| $V=\pi r^2 h = 355 \Rightarrow h=\frac{355}{\pi r^2}$ (or $rh=\frac{355}{\pi r}$ or $\pi rh=\frac{355}{r}$) | B1 | 1.1b — Correct expression for $h$ or $rh$ or $\pi rh$ in terms of $r$; may be implied by later substitution |
| $C=0.04(\pi r^2+2\pi rh)+0.09(\pi r^2)$ | M1 | 3.4 — Scored for sum of three terms of form $0.04...r^2$, $0.09...r^2$ and $0.04\times...rh$ |
| $C=0.13\pi r^2+0.08\pi rh=0.13\pi r^2+0.08\pi r\left(\frac{355}{\pi r^2}\right)$ | dM1 | 2.1 — Substitutes $h$ into $C$; dependent on correct form of $h$ |
| $C=0.13\pi r^2+\frac{28.4}{r}$ | A1* | 1.1b — Fully correct, no errors. Must see: separate volume equation, separate costs before combining, substitution before $\frac{28.4}{r}$ term |
**Part (b):**
| Working | Mark | Guidance |
|---------|------|----------|
| $\frac{dC}{dr}=0.26\pi r - \frac{28.4}{r^2}$ | M1, A1 | 3.4, 1.1b — M1: differentiates to get at least $r^{-1}\to r^{-2}$; A1: correct derivative |
| $\frac{dC}{dr}=0 \Rightarrow r^3=\frac{28.4}{0.26\pi} \Rightarrow r=...$ | M1 | 1.1b — Sets $\frac{dC}{dr}=0$, solves for $r$; do not withhold if $r$ is negative |
| $r=\sqrt[3]{\frac{1420}{13\pi}}=3.26...$ | A1 | 1.1b — Correct value; allow exact or awrt 3.26 |
**Part (c):**
| Working | Mark | Guidance |
|---------|------|----------|
| $\frac{d^2C}{dr^2}=0.26\pi+\frac{56.8}{r^3}=0.26\pi+\frac{56.8}{\text{"3.26"}^3}$ | M1 | 1.1b — Finds $\frac{d^2C}{dr^2}$ at their positive $r$; valid if second derivative is of form $A+\frac{B}{r^3}$ where $A,B\neq 0$ |
| $\frac{d^2C}{dr^2}=(2.45...)>0$, hence minimum (cost) | A1 | 2.4 — Requires correct $\frac{d^2C}{dr^2}$ AND deduction that $\frac{d^2C}{dr^2}>0$ with minimal explanation |
**Part (d):**
| Working | Mark | Guidance |
|---------|------|----------|
| $C=0.13\pi(\text{"3.26"})^2+\frac{28.4}{\text{"3.26"}}$ | M1 | 3.4 — Uses model with positive $r$ from (b); may be seen in (b) |
| $(C=)13$ | A1 | 1.1b — Ignore units |
---\begin{enumerate}
\item A company makes drinks containers out of metal.
\end{enumerate}
The containers are modelled as closed cylinders with base radius $r \mathrm {~cm}$ and height $h \mathrm {~cm}$ and the capacity of each container is $355 \mathrm {~cm} ^ { 3 }$
The metal used
\begin{itemize}
\item for the circular base and the curved side costs 0.04 pence/ $\mathrm { cm } ^ { 2 }$
\item for the circular top costs 0.09 pence/ $\mathrm { cm } ^ { 2 }$
\end{itemize}
Both metals used are of negligible thickness.\\
(a) Show that the total cost, $C$ pence, of the metal for one container is given by
$$C = 0.13 \pi r ^ { 2 } + \frac { 28.4 } { r }$$
(b) Use calculus to find the value of $r$ for which $C$ is a minimum, giving your answer to 3 significant figures.\\
(c) Using $\frac { \mathrm { d } ^ { 2 } C } { \mathrm {~d} r ^ { 2 } }$ prove that the cost is minimised for the value of $r$ found in part (b).\\
(d) Hence find the minimum value of $C$, giving your answer to the nearest integer.
\hfill \mbox{\textit{Edexcel AS Paper 1 2022 Q12 [12]}}