Edexcel C2 — Question 6 12 marks

Exam BoardEdexcel
ModuleC2 (Core Mathematics 2)
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicStationary points and optimisation
TypeShow formula then optimise: cylinder/prism (single variable)
DifficultyStandard +0.3 This is a standard C2 optimization problem with a volume constraint. Part (a) is straightforward substitution using V=πr²h=500. Parts (b-d) involve routine differentiation, solving dA/dr=0, and applying the second derivative test—all textbook techniques with no novel insight required. Slightly easier than average due to clear structure and standard methods.
Spec1.02z Models in context: use functions in modelling1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx

6. A container made from thin metal is in the shape of a right circular cylinder with height \(h \mathrm {~cm}\) and base radius \(r \mathrm {~cm}\). The container has no lid. When full of water, the container holds \(500 \mathrm {~cm} ^ { 3 }\) of water.
  1. Show that the exterior surface area, \(A \mathrm {~cm} ^ { 2 }\), of the container is given by \(A = \pi r ^ { 2 } + \frac { 1000 } { r }\).
  2. Find the value of \(r\) for which \(A\) is a minimum.
  3. Prove that this value of \(r\) gives a minimum value of \(A\).
  4. Calculate the minimum value of \(A\), giving your answer to the nearest integer.

Question 6:
Part (a):
AnswerMarks Guidance
\(V = \pi r^2 h = 500\), \(A = 2\pi r h + \pi r^2\)B1, M1
\(A = 2\pi r\left(\frac{500}{\pi r^2}\right) + \pi r^2 = \pi r^2 + \frac{1000}{r}\)M1 A1 (4 marks)
Part (b):
AnswerMarks Guidance
\(\frac{dA}{dr} = 2\pi r - 1000r^{-2}\)M1 A1
\(2\pi r - 1000r^{-2} = 0\), \(r = \sqrt[3]{\frac{500}{\pi}}\) (\(\approx 5.42\))M1 A1 (4 marks)
Part (c):
AnswerMarks Guidance
\(\frac{d^2A}{dr^2} = 2\pi + 2000r^{-3} > 0\), therefore minimumM1 A1 ft (2 marks)
Part (d):
AnswerMarks Guidance
\(A = \pi r^2 + \frac{1000}{r} = 277\) (nearest integer)M1 A1 (2 marks)
# Question 6:

## Part (a):
$V = \pi r^2 h = 500$, $A = 2\pi r h + \pi r^2$ | B1, M1 |
$A = 2\pi r\left(\frac{500}{\pi r^2}\right) + \pi r^2 = \pi r^2 + \frac{1000}{r}$ | M1 A1 | (4 marks)

## Part (b):
$\frac{dA}{dr} = 2\pi r - 1000r^{-2}$ | M1 A1 |
$2\pi r - 1000r^{-2} = 0$, $r = \sqrt[3]{\frac{500}{\pi}}$ ($\approx 5.42$) | M1 A1 | (4 marks)

## Part (c):
$\frac{d^2A}{dr^2} = 2\pi + 2000r^{-3} > 0$, therefore minimum | M1 A1 ft | (2 marks)

## Part (d):
$A = \pi r^2 + \frac{1000}{r} = 277$ (nearest integer) | M1 A1 | (2 marks)

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6. A container made from thin metal is in the shape of a right circular cylinder with height $h \mathrm {~cm}$ and base radius $r \mathrm {~cm}$. The container has no lid. When full of water, the container holds $500 \mathrm {~cm} ^ { 3 }$ of water.
\begin{enumerate}[label=(\alph*)]
\item Show that the exterior surface area, $A \mathrm {~cm} ^ { 2 }$, of the container is given by $A = \pi r ^ { 2 } + \frac { 1000 } { r }$.
\item Find the value of $r$ for which $A$ is a minimum.
\item Prove that this value of $r$ gives a minimum value of $A$.
\item Calculate the minimum value of $A$, giving your answer to the nearest integer.
\end{enumerate}

\hfill \mbox{\textit{Edexcel C2  Q6 [12]}}