Ratio volume of cylinder inside cone

[songoku]

Homework Statement

Show that the volume of an upright cylinder that can be inscribe in an upright cone is 4/9 times the volume of cone

Homework Equations

volume of cone
volume of cylinder
differentiation ??
similarity of triangle

The Attempt at a Solution

I draw the picture of cylinder inside a cone and by using similarity of triangle I got:
(H-h) / h = r / R
where: H = height of cone, h = height of cylinder, r = radius of cylinder, R = radius of cone

But I can't find the answer. The final form I can get is:
V_cylinder / V_cone = 3 (H - h)²/(Hh)

I encountered this problem in differentiation chapter. How to use differentiation to solve this problem?

Thanks

 

[lepton5]

yes, you are correct using similarity of triangle. After that you have to use the differentiation to optimize cylinder's volume, [tex]dV_c = 0[/tex].

Subsitute for h from the ratio into above eqn, to get the radius of cylinder.

and, finally you can proof it the ratio for cylinder's volume to cone.

 

[songoku]

yes, you are correct using similarity of triangle. After that you have to use the differentiation to optimize cylinder's volume, [tex]dV_c = 0[/tex].

Subsitute for h from the ratio into above eqn, to get the radius of cylinder.

and, finally you can proof it the ratio for cylinder's volume to cone.

From (H-h) / h = r / R ==> r = R (H-h) / h

V_cylinder=πr²h , substitute r from above equation
=πR²(H-h)² / h

Assume R and H are constant and differentiating with respect to h
dV / dh = 0
-2πR²(H-h)h - πR²(H-h)²=0

After a little work, -h = H ?

 

[lepton5]

It's more simple if you substitue for h, since h in eqn of cylinder volume is not quadratic form.

so subs [tex]h = H - \frac{H}{R} . r[/tex] to volume of cylinder.


then optimize it, the differential is more simple with this way.

 

[songoku]

It's more simple if you substitue for h, since h in eqn of cylinder volume is not quadratic form.

so subs [tex]h = H - \frac{H}{R} . r[/tex] to volume of cylinder.


then optimize it, the differential is more simple with this way.

I can't find the answer because the my equation obtained from similarity of triangle is different than yours. I don't know how to obtain your equation. Can you explain it a little bit more because from similarity I got (H-h) / h = r / R

Thanks

 

[lepton5]

use this one, [tex] \frac{H - h}{H} = \frac{r}{R}[/tex].


you are wrong when you use h as denominator at left side, since you have equate it with R (radius of cone) you must also use H (height of cone).

can you see : big triangle vs little triangle (from top of cone).

 

[songoku]

use this one, [tex] \frac{H - h}{H} = \frac{r}{R}[/tex].


you are wrong when you use h as denominator at left side, since you have equate it with R (radius of cone) you must also use H (height of cone).

can you see : big triangle vs little triangle (from top of cone).

Ahh, why don't I realize it. Thanks