Heres a calc problem.
I have to find the maximum volume of the upside down cone inside the larger cone.
Thanks
Hello
Here is a cut on a plane
Volume of "small" cone is v = 1/3 pi rē h
It is possible to express r and h function of theta, R and H
First
$\displaystyle h\;=\;\frac{r}{tan \theta}$
Then
$\displaystyle \frac{r}{tan \alpha} + \frac{r}{tan \theta}\;=\;H$
After simplifications
$\displaystyle r\;=\;\frac{RHtan \theta}{R+Htan \theta}$
Thus
$\displaystyle v\;=\;\frac{1}{3}\pi R^3 H^3 \frac{(tan \theta)^2}{(R+Htan \theta)^3}$
After differentiating I find an extremum for
$\displaystyle tan \theta\;=\;\frac{H+\sqrt{H^2+6RH}}{3H}$
With R=6' and H=12'
$\displaystyle tan \theta\;=\;1$
$\displaystyle \theta\;=\;45°$
Corresponding volume is
$\displaystyle v\;=\;\frac{1}{3}\pi\frac{R^3H^3}{(R+H)^3}$
which is about 15% of the volume of the "big" cone
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.
$\displaystyle \frac{H-h}{r}=\frac{H}{R}$
$\displaystyle r=\frac{(H-h)R}{H}$
Plug this into the cone volume formula:
$\displaystyle V=\frac{\pi}{3}\left(\frac{(H-h)R}{H}\right)^{2}\cdot h$
Now differentiate w.r,t h, set to 0 and solve for h. H=12, R=6 are constants
You should get $\displaystyle h=\frac{H}{3}$ and go from there.