CAIE P1 2002 June — Question 8 10 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2002
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicChain Rule
TypeOptimization with constraint
DifficultyStandard +0.3 This is a standard optimization problem requiring constraint manipulation, differentiation of a polynomial, and second derivative test. The algebra is straightforward, and the problem follows a familiar textbook pattern with clear scaffolding through parts (i)-(iii). Slightly above average due to the multi-step nature, but well within typical P1 expectations.
Spec1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx1.07p Points of inflection: using second derivative

8 A hollow circular cylinder, open at one end, is constructed of thin sheet metal. The total external surface area of the cylinder is \(192 \pi \mathrm {~cm} ^ { 2 }\). The cylinder has a radius of \(r \mathrm {~cm}\) and a height of \(h \mathrm {~cm}\).
  1. Express \(h\) in terms of \(r\) and show that the volume, \(V \mathrm {~cm} ^ { 3 }\), of the cylinder is given by $$V = \frac { 1 } { 2 } \pi \left( 192 r - r ^ { 3 } \right) .$$ Given that \(r\) can vary,
  2. find the value of \(r\) for which \(V\) has a stationary value,
  3. find this stationary value and determine whether it is a maximum or a minimum.

(i) \(192\pi - \pi r^2 + 2\pi rh\)
leads to \(h = (192\pi - \pi r^2) \div 2\pi r\)
AnswerMarks Guidance
\(V = \pi r^2 h = \frac{1}{2}\pi(192r - r^3)\)M1, A1, M1, A1 Tries to relate surface area and (1 or 2) circles. Correct only. Subs for h into a correct volume formula. Answer was given (beware fortuitous ans)
(ii) \(\frac{dV}{dr} = \frac{1}{2}\pi(192 - 3r^2) = 0\) when \(r = 8\)M1, DM1, A1 Attempt to differentiate. Attempt to set to 0. Correct only
(iii) Value of \(V = 1610\) (or \(512\pi\))A1 Correct only – could be in (ii)
\(\frac{d^2V}{dr^2} = \frac{1}{2}\pi(-6r)\) Negative
AnswerMarks Guidance
maximumM1, A1\*, 3 Any correct method for max/min. Correct conclusion (must have second differential correct, but for "+" is ok)
**(i)** $192\pi - \pi r^2 + 2\pi rh$
leads to $h = (192\pi - \pi r^2) \div 2\pi r$
$V = \pi r^2 h = \frac{1}{2}\pi(192r - r^3)$ | M1, A1, M1, A1 | Tries to relate surface area and (1 or 2) circles. Correct only. Subs for h into a correct volume formula. Answer was given (beware fortuitous ans)

**(ii)** $\frac{dV}{dr} = \frac{1}{2}\pi(192 - 3r^2) = 0$ when $r = 8$ | M1, DM1, A1 | Attempt to differentiate. Attempt to set to 0. Correct only

**(iii)** Value of $V = 1610$ (or $512\pi$) | A1 | Correct only – could be in (ii)

$\frac{d^2V}{dr^2} = \frac{1}{2}\pi(-6r)$ Negative
maximum | M1, A1\*, 3 | Any correct method for max/min. Correct conclusion (must have second differential correct, but for "+" is ok)
8 A hollow circular cylinder, open at one end, is constructed of thin sheet metal. The total external surface area of the cylinder is $192 \pi \mathrm {~cm} ^ { 2 }$. The cylinder has a radius of $r \mathrm {~cm}$ and a height of $h \mathrm {~cm}$.\\
(i) Express $h$ in terms of $r$ and show that the volume, $V \mathrm {~cm} ^ { 3 }$, of the cylinder is given by

$$V = \frac { 1 } { 2 } \pi \left( 192 r - r ^ { 3 } \right) .$$

Given that $r$ can vary,\\
(ii) find the value of $r$ for which $V$ has a stationary value,\\
(iii) find this stationary value and determine whether it is a maximum or a minimum.

\hfill \mbox{\textit{CAIE P1 2002 Q8 [10]}}