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3) Calculate the dimensions of the straight circular cone, smaller volume that can be circumscribed

leprofece

Member
Jan 23, 2014
241
3) Calculate the dimensions of the straight circular cone, smaller volume that can be circumscribed around a cylinder of RADIUS "R" and height "H".

Answer is h = 3H and r= 3R/2
 
Last edited:

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,506
First, see rule #11 here (select "Explained Rules" instead of "General Rules" in the dropbox at the top of the page for more information).

3) Calculate the dimensions of the straight circular cone, smaller volume that can be circumscribed around a cylinder of RADIUS "R" and height "H".
I assume you need a cone with the smallest volume. I recommend denoting the dimensions (radius and height) of the cylinder by $r$ and $h$ and those of the cone by $R$ and $H$, since they are bigger. You can parametrize the problem by $R$ and express $H$ through $R$, $r$ and $h$. After that, express the volume of the cone and minimize it.
 

leprofece

Member
Jan 23, 2014
241
First, see rule #11 here (select "Explained Rules" instead of "General Rules" in the dropbox at the top of the page for more information).

I assume you need a cone with the smallest volume. I recommend denoting the dimensions (radius and height) of the cylinder by $r$ and $h$ and those of the cone by $R$ and $H$, since they are bigger. You can parametrize the problem by $R$ and express $H$ through $R$, $r$ and $h$. After that, express the volume of the cone and minimize it.
I need to show the answer provided Could you give the entire procedure???
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
We need *you* to show some work...what have you done so far?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I need to show the answer provided Could you give the entire procedure???
Hello Jose,

Our primary goal here at MHB is to assist students in coming to a solution themselves. You will benefit much more in actively participating in the problem versus watching someone else work the problem. The best way we can do this is if you show your work or thoughts and this will give us a better idea where you are stuck and need assistance.

Most times, it is very beneficial to have it explained to us why what we are trying is wrong. You may very well not realize this just seeing someone else work the problem, and then you may make the same type of mistake on a future problem, without knowing why it is wrong.

This is why we expect effort to be shown, because not only does it make it easier for our helpers to address specific issues you may have, you then build more confidence to do other problems in the future as you have already taken an active role in this problem.

So, can you show how you would apply the help that has already been given?
 

leprofece

Member
Jan 23, 2014
241
Hello Jose,

Our primary goal here at MHB is to assist students in coming to a solution themselves. You will benefit much more in actively participating in the problem versus watching someone else work the problem. The best way we can do this is if you show your work or thoughts and this will give us a better idea where you are stuck and need assistance.

Most times, it is very beneficial to have it explained to us why what we are trying is wrong. You may very well not realize this just seeing someone else work the problem, and then you may make the same type of mistake on a future problem, without knowing why it is wrong.

This is why we expect effort to be shown, because not only does it make it easier for our helpers to address specific issues you may have, you then build more confidence to do other problems in the future as you have already taken an active role in this problem.

So, can you show how you would apply the help that has already been given?
ok Volume cone V = pi r2h from there on , I can not apply the pitagorean calculus to get
the cone into the cylinder
something like that
h/r = h-r/R
as it is suggested in my book
It is a very difficult problem like the others what I have asked noy very commons in books and universities but here in Venezuela they are asked in a very good private university
I appreciate your help
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
This is what I would do:

Draw a cross-section of the cylinder and cone through its apex (and the widest part of the cylinder) perpendicular to the base of the cone and cylinder.

If the cylinder has height H, and radius R, draw the origin at the center of the base (so it is at (0,0)). Then we know the edge of the cone and the top of the cylinder intersect at the point (R,H).

It should be clear the apex of the cone is at (0,h) and the widest part of the base is at (r,0).

Determine a formula for the line that goes through (0,h), (R,H) and (r,0). This will let you deduce what h is in terms of r,R, and H. My suggestion is to use the "two-point" formula for a line, using (R,H) and (r,0) as your 2 points.

Next, set $V(r) = \pi r^2h$.

Substitute your value for h in as a formula in r,R and H (R and H are constants, so we get a function of just one variable, r).

Use some property you learned from calculus about when V has a local maximum or minimum.

If you can find either of r and h, finding the other one should be easy.
 

leprofece

Member
Jan 23, 2014
241
a line well thanks for your trying but i think you are getting more confused to me
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,506
It should be clear the apex of the cone is at (0,h) and the widest part of the base is at (r,0).
According to which definitions of $r$ and $h$?
I recommend denoting the dimensions (radius and height) of the cylinder by $r$ and $h$ and those of the cone by $R$ and $H$, since they are bigger.
Determine a formula for the line that goes through (0,h), (R,H) and (r,0). This will let you deduce what h is in terms of r,R, and H. My suggestion is to use the "two-point" formula for a line, using (R,H) and (r,0) as your 2 points.
Alternatively, use the similarity of triangles in the cross-section.

Next, set $V(r) = \pi r^2h$.
Should be $(1/3)\pi r^2h$, but the constant multiplier is irrelevant for finding the point of minimum.
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,506
Volume cone V = pi r2h
Times 1/3.

I can not apply the pitagorean calculus to get
the cone into the cylinder
You don't need the Pythagorean theorem.

something like that
h/r = h-r/R
as it is suggested in my book
You can get something like that. Let the dimensions of the cylinder be $r,h$ and the dimensions of the cone be $R,H$. Then from similar triangles we get
\[
\frac{h}{R-r}=\frac{H}{R}, \text{ or }H=\frac{hR}{R-r}
\]
Thus, the volume of the cone is
\[
V(R)=\frac{\pi hR^3}{3(R-r)}
\]
The constant factor $\pi h/3$ does not change the $R$ for which the expression is minimized. We have
\[
\frac{d}{dR}\left(\frac{R^3}{R-r}\right)=\frac{3R^2(R-r)-R^3}{(R-r)^2}
\]
Setting the denominator numerator to zero gives the requires expression of $R$ through $r$.
 
Last edited:

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
Yes I forgot about the 1/3. It turns out it won't matter.

Anyway, what I meant is:

The line that is rotated about the y-axis to form the cone is:

$y = \dfrac{0 - H}{r - R}(x - R) + H$

I am using the original poster's H and R, they're just SYMBOLS and suggesting he change them to suit what you think they ought to be seems to add extra confusion.

since (0,h) lies on this line, we get that:

$h = \dfrac{-H}{r - R}(-R) + H$

Putting everything over a common denominator gives:

$h = \dfrac{HR + Hr - HR}{r - R} = \dfrac{Hr}{r - R}$

so we have:

$V(r) = \dfrac{\pi Hr^3}{3(r - R)}$

The factor $\dfrac{\pi H}{3}$ is just a constant, we need to differentiate:

$f(r) = \dfrac{r^3}{r - R}$.

The quotient rule gives:

$f'(r) = \dfrac{(r - R)(3r^2) - (r^3)}{(r - R)^2} = \dfrac{r^2(3(r - R) - r)}{(r - R)^2}$

$= \dfrac{r^2(2r - 3R)}{(r - R)^2}$

which is 0 when the denominator is 0.

One of the possible solutions can be discarded, since it reflects a null cone.