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

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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) About Me  (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 H and h ; their radii are R and r .
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:
r%2FR=%28H-h%29%2FH <--> r%2FR=1-h%2FH <-- r=R%281-h%2FH%29 .
The volume of the small cone is
v=pi%2Ar%5E2h%2F3=%28pi%2F3%29r%5E2h .
Substituting R%281-h%2FH%29 for r , we get
v=%28pi%2F3%29%28R%281-h%2FH%29%29%5E2h
v=%28pi%2F3%29R%5E2%281-h%2FH%29%5E2h
v=%28pi%2F3%29R%5E2%281-2h%2FH%2Bh%5E2%2FH%5E2%29h
v=%28pi%2F3%29R%5E2%28h-2h%5E2%2FH%2Bh%5E3%2FH%5E2%29
That is a function of h, with r and H being constants.
The derivative is
dv%2Fdr=%28pi%2F3%29R%5E2%281%2B4h%2FH%2B3h%5E2%2FH%5E2%29
The maxima and minima of v will happen when dv%2Fdr=0 ,
and that will happen when 1%2B4h%2FH%2B3h%5E2%2FH%5E2=0 .
If you change the name of the variables to x=h%2FH , you can re-write the equation as
3x%5E2-4x%2B1=0 , and you would recognize it as a quadratic equation,
with solutions x=1%2F3(meaning highlight%28h%2FH=1%2F3%29 )
and x=1 (meaning h%2FH=1 ).
The polynomial 1%2B4h%2FH%2B3h%5E2%2FH%5E2=0 , and dv%2Fdr changes sign at each of its two zeros,
so one must be a maximum and the other a minimum.
h%2FH=1 makes r=0 and v=0 , so for h%2FH=1 the small cone volume is minimum.
The maximum volume for the small cone happens when highlight%28h%2FH=1%2F3%29 .