# Maximum volume of cone

• Jun 13th 2011, 12:03 AM
Punch
Maximum volume of cone
A right-angled triangle $\displaystyle ABC$ with hypotenuse $\displaystyle AB$ of length $\displaystyle 4$metres is revolved about $\displaystyle BC$ to generate a right circular cone. Show that the maximum volume of the cone is obtained when the ratio $\displaystyle BC:CA$ is $\displaystyle 1:\sqrt{2}$
• Jun 13th 2011, 12:24 AM
Unknown008
Let h = BC and r = AC

You get the equation: $\displaystyle h^2 + r^2 = 4^2$

The volume of a cone is then:

$\displaystyle V = \frac13\pi r^2 h$

Max volume is given by the derivative of the volume and set to 0. You have to use a substitution so that you have the volume in terms of r only, or in terms of h only.

You will have a value of r or h, depending on what you did and you can substitute that in the first equation to get the value of the other unknown, after which, you should get the ratio asked for.

EDIT: Really silly mistakes (Doh)
• Jun 13th 2011, 01:21 AM
Punch
Quote:

Originally Posted by Unknown008
Let r = BC and h = AC

You get the equation: $\displaystyle h^2 + r^2 = 4$

The volume of a cone is then:

$\displaystyle V = \frac23\pi r^2 h$

Max volume is given by the derivative of the volume and set to 0. You have to use a substitution so that you have the volume in terms of r only, or in terms of h only.

You will have a value of r or h, depending on what you did and you can substitute that in the first equation to get the value of the other unknown, after which, you should get the ratio asked for.

Let $\displaystyle BC$ be $\displaystyle r$ and $\displaystyle AC$ be $\displaystyle h$
$\displaystyle V=\frac{1}{3}{\pi}r^2h$

$\displaystyle r^2+h^2=4^2$
$\displaystyle r^2=16-h^2$

$\displaystyle V=\frac{1}{3}{\pi}(16-h^2)h$
$\displaystyle \frac{dV}{dh}=\frac{h}{3}{\pi}(-2h)+(16-h^2)(\frac{1}{3}{\pi})$

$\displaystyle \frac{dV}{dh}=0$,

$\displaystyle \frac{h}{3}{\pi}(-2h)+(16-h^2)(\frac{1}{3}{\pi})=0$
$\displaystyle -\frac{2}{3}{\pi}h^2+\frac{16}{3}{\pi}-\frac{h^2}{3}{\pi}=0$
$\displaystyle (\frac{2}{3}{\pi}+\frac{{\pi}}{3})h^2=\frac{16}{3} {\pi}$
$\displaystyle h=2.309$

sub $\displaystyle h=2.309, r=3.266$

$\displaystyle \frac{r}{h}=BC:CA=1:1.414=1:\sqrt{2}$ (shown)
• Jun 13th 2011, 01:40 AM
Unknown008
Oops, I just realised I gave you the wrong formula for the volume and the pythagoras' theorem. And... BC was supposed to be h and AC be r. My bad, I'm losing my touch (Crying)

You could have kept it in exact form here:

$\displaystyle \left(\frac23 \pi + \frac{\pi}{3}\right)h^2 = \frac{16}{3}\pi$

$\displaystyle \pi h^2 = \frac{16}{3}\pi$

$\displaystyle h^2 = \frac{16}{3}$

$\displaystyle h = \frac{4}{\sqrt{3}}$

Then $\displaystyle r = \sqrt{\frac{48-16}{3}} = \frac{4\sqrt2}{\sqrt3}$

Then the ratio = AC:BC = $\displaystyle \frac{4\sqrt2}{\sqrt3} : \frac{4}{\sqrt{3}} = \sqrt2 : 1$