19M.1.AHL.TZ2.H_8

pestleMathematicsAIHLPaper 119M· sl-5-6-stationary-points-local-max-and-minsource ↗

A right circular cone of radius $r$ is inscribed in a sphere with centre O and radius $R$ as shown in the following diagram. The perpendicular height of the cone is $h$ , X denotes the centre of its base and B a point where the cone touches the sphere.

Show that the volume of the cone may be expressed by $V = \frac{\pi }{3}\left( {2R{h^2} - {h^3}} \right)$ .

[4]

Given that there is one inscribed cone having a maximum volume, show that the volume of this cone is $\frac{{32\pi {R^3}}}{{81}}$ .

[4]

Markscheme / solution

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

attempt to use Pythagoras in triangle OXB M1

$\Rightarrow {r^2} = {R^2} - {\left( {h - R} \right)^2}$ A1

substitution of their ${r^2}$ into formula for volume of cone $V = \frac{{\pi {r^2}h}}{3}$ M1

$= \frac{{\pi h}}{3}\left( {{R^2} - {{\left( {h - R} \right)}^2}} \right)$

$= \frac{{\pi h}}{3}\left( {{R^2} - \left( {{h^2} + {R^2} - 2hR} \right)} \right)$ A1

Note: This A mark is independent and may be seen anywhere for the correct expansion of ${{{\left( {h - R} \right)}^2}}$ .

$= \frac{{\pi h}}{3}\left( {2hR - {h^2}} \right)$

$= \frac{\pi }{3}\left( {2R{h^2} - {h^3}} \right)$ AG

[4 marks]

at max, $\frac{{{\text{d}}V}}{{{\text{d}}h}} = 0$ R1

$\frac{{{\text{d}}V}}{{{\text{d}}h}} = \frac{\pi }{3}\left( {4Rh - 3{h^2}} \right)$

$\Rightarrow 4Rh = 3{h^2}$

$\Rightarrow h = \frac{{4R}}{3}$ (since $h \ne 0$ ) A1

EITHER

${V_{{\text{max}}}} = \frac{\pi }{3}\left( {2R{h^2} - {h^3}} \right)$ from part (a)

$= \frac{\pi }{3}\left( {2R{{\left( {\frac{{4R}}{3}} \right)}^2} - {{\left( {\frac{{4R}}{3}} \right)}^3}} \right)$ A1

$= \frac{\pi }{3}\left( {2R\frac{{16{R^2}}}{9} - \left( {\frac{{64{R^3}}}{{27}}} \right)} \right)$ A1

${r^2} = {R^2} - {\left( {\frac{{4R}}{3} - R} \right)^2}$

${r^2} = {R^2} - \frac{{{R^2}}}{9} = \frac{{8{R^2}}}{9}$ A1

$\Rightarrow {V_{{\text{max}}}} = \frac{{\pi {r^2}}}{3}\left( {\frac{{4R}}{3}} \right)$

$= \frac{{4\pi R}}{9}\left( {\frac{{8{R^2}}}{9}} \right)$ A1

THEN

$= \frac{{32\pi {R^3}}}{{81}}$ AG

[4 marks]

Examiners’ report

[N/A]