1. ## Cone Optimization

Heres a calc problem.

I have to find the maximum volume of the upside down cone inside the larger cone.

Thanks

2. Hello

Here is a cut on a plane

Volume of "small" cone is v = 1/3 pi r² h
It is possible to express r and h function of theta, R and H
First
$h\;=\;\frac{r}{tan \theta}$

Then
$\frac{r}{tan \alpha} + \frac{r}{tan \theta}\;=\;H$

After simplifications
$r\;=\;\frac{RHtan \theta}{R+Htan \theta}$

Thus
$v\;=\;\frac{1}{3}\pi R^3 H^3 \frac{(tan \theta)^2}{(R+Htan \theta)^3}$

After differentiating I find an extremum for
$tan \theta\;=\;\frac{H+\sqrt{H^2+6RH}}{3H}$

With R=6' and H=12'

$tan \theta\;=\;1$

$\theta\;=\;45°$

Corresponding volume is
$v\;=\;\frac{1}{3}\pi\frac{R^3H^3}{(R+H)^3}$

which is about 15% of the volume of the "big" cone

3. This is a little different than the usual cylinder in a cone and what not.

But, like most others we can use similar triangles. Here it is in general. You can add the constants H=12 and R=6

Let H=height of big cone, R=radius of big cone, r=radius of small cone, h=height of small cone.

$\frac{H-h}{r}=\frac{H}{R}$

$r=\frac{(H-h)R}{H}$

Plug this into the cone volume formula:

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

Now differentiate w.r,t h, set to 0 and solve for h. H=12, R=6 are constants

You should get $h=\frac{H}{3}$ and go from there.

4. Much more simple than mine !

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### cone optimization

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