Solving for radius of a combined shape of a cone and a cylinder where the cone is base is concentric with the cylinder? I have a solid that is a combined shape of a cylinder and a concentric cone
(a round sharpened pencil would be a good example)
Know values are:
Total Volume = 46,000
Height to Base Ratio =  2/1
(Height = cone height + cylinder height) (Base = Diameter)
Angle of cone slope  =  30 degrees (between base and slope of cone)
How do you solve for the Radius?
2  let say the ratio is from the orignal length 
Know value:
Total Volume = V = 46,000
Original Height to Base Ratio = t =  2/1
(Height = cone height + cylinder height + distance shortened) (Base = Diameter)
Angle of cone slope  = θ =  30 degrees (between base and slope of cone)
distance shortened from original Height x = 3 
(Let h be the height of the cylinder. Then h+rtanθ+x=2(2r)
How do you solve for the Radius?
 A: I assume that by height to base ratio you mean the ratio of total height (cylinder plus cone) to the diameter of the base.
Let $\theta$ be the angle between the base of the cone and the sloping walls. Then the height of the cone is $r\tan\theta$. In our case $\tan\theta=\frac{1}{\sqrt{3}}$. Kind of a stubby cone. We keep on writing $\tan\theta$ instead of $\frac{1}{\sqrt{3}}$, it makes typing easier.
Let $h$ be the height of the cylinder. Then $h+r\tan\theta=2(2r)$, and therefore $h=r(4-\tan\theta)$.
The combined volume is $\pi r^2h+\frac{1}{3}\pi r^3\tan\theta$. Substituting for $h$, and simplifying a little, we find that the volume is
$$\pi\left(4-\frac{2}{3}\tan\theta \right)r^3.$$
Now we can set this equal to $46000$ and solve for $r$. I get something close to $15.94$, but do check!
A: Cone height = $ \sqrt 3 R$. 
Cylinder height = Total height - Cone heght 
$$= (4 -\sqrt3 R ) $$
Total Volume 
$$ V_{total} = V_{cone}+ V_{cyl  } =  \pi R^2 [ (4- \sqrt 3) R ] + \frac\pi3 R^2 \cdot \sqrt3 R = 46000. $$
from  which only unknown $R^3, R $ can be calculated.
