[Guest]

A sphere of radius r is inside a cone as shown in the figure. If the cone height h is made larger, then the radius of the base b needs to be shorter in order for the sphere to be in contact with the sides and the bottom of the cone.

Find h (as a function of r) That minimizes the volume of the cone. In other words find the the height of the cone with minimum volume that satisfies the conditions shown in the figure.

P.S. This if my first post. English is my second language and I am Engineer not a mathematician.

Edited April 2, 2009 by momo567

[bonanova]

h=3r.

Spoiler for and:

Cone angle is 60^o.

Your English is great.

Nice puzzle.

[bonanova]

Reduce to 2 dimensions and think of the area of a triangle

[which is rotated about y-axis to form volume of cone.]

Center a unit radius circle at the origin.

From a point y on the y-axis drop a tangent that intersects the line y = -1 at the value x.

The line makes an angle theta with the y-axis.

sin theta = 1/y --> y+1 = [1+sin theta]/sin theta

tan theta = x/[y+1] --> x = [y+1] tan theta = [1+sin theta]/cos theta

Area of triangle [which if minimized will minimize volume of cone] = x[y+1]/2

Area = [1/2] {[1+sin theta]/sin theta} {[1 + sin theta]/cos theta} = ... = [1 + sin theta]²/sin 2theta

Do the messy differention, set to zero and get theta = 30^o.

Now consider the drawing:

Sin theta = r/[h-r] = .5 --> 2r = h-r ==> h = 3r.

The cone is half again as high as the diameter of the sphere.

[Guest]

Your English is great.

Nice puzzle.

h=3r.

Spoiler for and:

Cone angle is 60^o.

Incorrect. The solution is not the same for the 2D counterpart of the problem (circle in triangle)

[Guest]

Reduce to 2 dimensions and think of the area of a triangle

[which is rotated about y-axis to form volume of cone.]

Center a unit radius circle at the origin.

From a point y on the y-axis drop a tangent that intersects the line y = -1 at the value x.

The line makes an angle theta with the y-axis.

sin theta = 1/y --> y+1 = [1+sin theta]/sin theta

tan theta = x/[y+1] --> x = [y+1] tan theta = [1+sin theta]/cos theta

Area of triangle [which if minimized will minimize volume of cone] = x[y+1]/2

Area = [1/2] {[1+sin theta]/sin theta} {[1 + sin theta]/cos theta} = ... = [1 + sin theta]²/sin 2theta

Do the messy differention, set to zero and get theta = 30^o.

Now consider the drawing:

Sin theta = r/[h-r] = .5 --> 2r = h-r ==> h = 3r.

The cone is half again as high as the diameter of the sphere.

Nice solution. But there is a flaw. The solution to the 2D and 3D problems is different.

[Guest]

Equation1: X1^2 + r ^2 = (h-r)^2

Equation2: h^2 + b^2 = (X1+b)^2

Combine to find equation 3: b=h*r/(sqrt(h^2- 2*h*r))

Volume of Cone Vc= (pi/3)* b^2 *h

Express as function of h. Differentiate and make equal to 0. Correct answer is....h=4

Edited April 2, 2009 by bonanova
Placed your figure inside the spoiler ;-)

[bonanova]

Ah... I can see now how 2D and 3D can be different.

Area sweeps out volume, but the parts of the triangle

at greater distance from y-axis generate more volume.

So you want a taller, thinner cone for the 3D case.

Nice analysis....

[bonanova]

You draw such nice diagrams, you might want to host a discussion with

of this related [sphere-in-a-cone] classic:

Given a wine glass [assume it's an inverted cone] with angle theta and known Height,

determine the Radius of the sphere which when inserted into the full glass will

displace the maximum volume of wine.

Clearly for a very small radius, the sphere can sink to the bottom of the glass

and displace its own small volume of wine.

And for very large radius, bottom surface of the sphere will approximate a

plane and not enter the glass at all, displacing a diminishingly small volume

of wine.

Somewhere in between, a maximum volume will be displaced.

Clearly the answer will be Radius = f[theta] x Height.

Determine f[theta].

[Guest]

Interesting problem. The diagrams look so nice because the program that I use allows me to use equations. I think at the end of the day I can take some time to figure out all the equations of the problem and draw the necessary figures. I'll let you know when I post the new puzzle.

Thanks!

[Guest]

dude just differentiate the volume of the cone wrt "r" and minimize it and u will get the answer...

no need for sin theta or cos theta....