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- Thread starter leprofece
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- Jan 30, 2012

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I assume you need a cone with the3) Calculate the dimensions of the straight circular cone, smaller volume that can be circumscribed around a cylinder of RADIUS "R" and height "H".

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I need to show the answer provided Could you give the entire procedure???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 thesmallestvolume. 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.

- Feb 15, 2012

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We need *you* to show some work...what have you done so far?

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Hello Jose,I need to show the answer provided Could you give the entire procedure???

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?

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ok Volume cone V = pi r2h from there on , I can not apply the pitagorean calculus to getHello 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?

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

- Feb 15, 2012

- 1,967

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.

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- Jan 30, 2012

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According to which definitions of $r$ and $h$?It should be clear the apex of the cone is at (0,h) and the widest part of the base is at (r,0).

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.

Alternatively, use the similarity of triangles in the cross-section.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.

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

- Jan 30, 2012

- 2,528

Times 1/3.Volume cone V = pi r^{2}h

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

the cone into the cylinder

Yousomething like that

h/r = h-r/R

as it is suggested in my book

\[

\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

Last edited:

- Feb 15, 2012

- 1,967

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.