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

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

* 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

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

$\Rightarrow 4Rh=3h^2$

$\Rightarrow h=\frac{4R}{3}$ (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

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

OR

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

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

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

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

THEN

$=\frac{32\pi R^3}{81}$       AG

[4 marks]