SOLUTION: A right circular cone is inscribed inside a larger right circular cone with a volume of 150 cm3. The axes of the cones coincide and the vertex of the inner cones touches the cente

Question 1030182: A right circular cone is inscribed inside a larger right circular cone with a volume of 150 cm3. The axes of the cones coincide and the vertex of the inner cones touches the center of the base of the outer cone. Find the ratio of the heights of the cones that maximizes the volume of the inner cone.
Answer by KMST(5328) (Show Source): You can put this solution on YOUR website!
By not giving enough data to find the radius of height, or height to radius ratio of the larger cone, the problem's wording hints that the answer does not depend on the height to radius ratio of the larger cone.
I do not believe you need to know the volume of the larger cone, either.

If there is a way to solve the problem without invoking calculus,
or if there is an easier or more elegant way to reach the answer,
let me know.
The way I saw to the solution is shown below.

Here is a cross-section of the cones, sliced along their axes.
The cones heights are and ; their radii are and .
The portion of the large cone above the base of the small cone is
a cone similar to the large cone (same shape, but scaled down).
Therefore, their corresponding length measurements are proportional:
<--> <-- .
The volume of the small cone is
.
Substituting for , we get

That is a function of , with and being constants.
The derivative is

The maxima and minima of will happen when ,
and that will happen when .
If you change the name of the variables to , you can re-write the equation as
, and you would recognize it as a quadratic equation,
with solutions (meaning )
and (meaning ).
The polynomial , and changes sign at each of its two zeros,
so one must be a maximum and the other a minimum.
makes and , so for the small cone volume is minimum.
The maximum volume for the small cone happens when .

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