Maximum volume of cone

**Punch** · June 13th 2011, 12:03 AM

A right-angled triangle $ABC$ with hypotenuse $AB$ of length $4$ metres is revolved about $BC$ to generate a right circular cone. Show that the maximum volume of the cone is obtained when the ratio $BC:CA$ is $1:\sqrt{2}$

**Unknown008** · June 13th 2011, 12:24 AM

Let h = BC and r = AC

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

The volume of a cone is then:

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

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

**Punch** · June 13th 2011, 01:21 AM

Originally Posted by Unknown008

Let r = BC and h = AC

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

The volume of a cone is then:

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

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

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

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

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

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

sub $h=2.309, r=3.266$

$\frac{r}{h}=BC:CA=1:1.414=1:\sqrt{2}$ (shown)

**Unknown008** · June 13th 2011, 01:40 AM

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

You could have kept it in exact form here:

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

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

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

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

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

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

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