Circles and Cones

[BobbyJoeCool]

To a bunch of Math people, this question will be pretty easy... But do it completly alebraicly (no graphing). We did this question in high school calculus, but we were able to use a graph. I refused to "cheat" by graphing it, and tried it the algebraic way. It took me several pieces of paper, but I was able to do it, and was very proud of myself, but I was trying to show someone how to do this and I forgot how I did it (I keep comming up with a whole bunch of Ø's and 2?'s that I couldn't get rid of).

 

You have a circle with radius "r." You cut out a sector of the circle at an angle of "Ø." You turn the two circle segments into cones by taking the two lines created by where the sector is pulled out, and put them together (if you can't visualize this, take a piece of paper, cut out a circle, then cut the radius of the circle at two points around the outside, thus cutting out a sector (like cutting a piece of pie). You fold the circle together so that the two lines of the radii you cut are touching and thus have created a cone).

 

What value of Ø will maximize the volume of the two cones?

 

I used to know the answer, but I'm more interested in the algebraic way of getting to the answer. Anyonw who knows how to do this, help is appreciated.. and those of you who what to try and figure it out just for fun, go right ahead.

 

Thanks.

[matt grime]

what do you mean maximize the volume of the "two cones"? the sum of the volumes? the product of the volumes?

[BobbyJoeCool]

I mean, the sum of the values of the two cones. "Maximize the total volume of the two cones." was the exact question if I remember correctly.

[Pat Says]

Do you mean all algebraicly or just no graphing. Because you could create a formula and find the max using calculus (no actual graphing involved). And I'm just guessing but wouldn't it always come out to be half of the circle?

[BobbyJoeCool]

Do you mean all algebraicly or just no graphing. Because you could create a formula and find the max using calculus (no actual graphing involved). And I'm just guessing but wouldn't it always come out to be half of the circle?

 

The idea is that, no matter what the radius of the circle, the angle Ø is the deciding factor in the volume of the cones (the bigger the radius, the bigger the volumes... but the angle of the sector cut out substancially changes the shape of the cones and that the volume for any given "r" can be maximized by finding the correct Ø. For every radius of the circle, the volume will be maximized by the same Ø. What is that Ø?)

 

There comes a point that you can find the answer by graphing and finding the maximum on the graph. What I'm saying is, don't do that.

[Pat Says]

I guess what I was asking is that you don't have to physically graph to find the max (can find where the rate = 0 ) so does that count? But, it sounds like you don't want to use rates. Otherwise, can't say I know how to do it. Other than I guess theta would equal pi (radians). Can you maybe show somewhere along the lines of what you did to give a hint or something?

[matt grime]

right, that's better, so you want to maximize the sum of the two volumes. (saying maximise two things does not a priori have meaning, is (1,3) bigger than (3,1) or (2,2)? if you don't state the ordering it is meaningless).

 

simple calculus will give you the answer, though symmetry makes the symmetric answer the favourite, but that is just a gut reaction and may turn out to be incorrect.

[Pat Says]

Yeah, I dunno exactly know what he means by algebra alone. Becuase to find the max you don't have to graph it.

[BobbyJoeCool]

first of all... you have to realize that for any radius of the original circle, the value of Ø*will be the same to maxamize... so we can assume a value for the radius (to make to easy, r=1).

 

As for the hint, I'll show you how to get the volumes of the cones....

 

what is the equation for the volume of a cone...

 

1/3*B*h.

 

B=2*?*r' (note, this is a different r)

 

the new r is found by taking the circumfrance of the base (2?*1-Ø (in radians)) and with the new r (r').

 

the radius of the original circle is also the slant height of the cone. (if you actually cut out the circle, you'll see why this is true, and you have the radius

 

the radius and slant height, and the height form a right triangle with the slant height as the hypotenuse. so

 

r^2=r'^2+h^2

r=1 and r'=2?-Ø so:

1=2?-Ø+h^2 so:

h^2=1-2?+Ø

so h=?(1-2?+Ø)

 

V1=1/3*(?*r'^2)*(?(1-(2?-Ø)^2))

V1=1/3*(?*(2?-Ø)^2)*(?(1-(2?-Ø)^2))

 

as for the other cone... "s" is the same, r'=Ø (in radians)

 

so

 

lets get h again..

 

h^2=r^2-r'^2

 

r=1, r'=Ø

 

h^2=1^2-Ø^2

 

V2=1/3*(?*(Ø)^2)*(?(1-Ø^2))

 

V1+V2=1/3*(?*(2?-Ø)^2)*(?(1-(2?-Ø)^2))+1/3*(?*(Ø)^2)*(?(1-Ø^2))

 

Maximize this equation...

[matt grime]

as i told you, you can do this by elementary, though tedious, analysis, which is "algebraic" without being graphical, though there is no harm in using graphical means since it is just finding the turning points, and if the answer happens to be the symmetric one so much the better i suppose. it is a question of when your patience tuns out that is all.

[BobbyJoeCool]

V1+V2=1/3*(?*(2?-Ø)^2)*(?(1-(2?-Ø)^2))+1/3*(?*(Ø)^2)*(?(1-Ø^2))

 

I tried to take the derivative of this equation and it proved very problematic (i did not have a calulator.) so that I could find the 0's at the derivative (turning points in the original equation). One of these is the maximum... the other is the minimum.

 

and just... something about this equation looks wrong as to what I did...

[Pat Says]

Yeah, it takes effing forever to find the derivative and then to solve for zero by hand but it can be done. The answer actually does turn out to be pi though so it is symmetrical. Arg solving for zero manually would be very tedious... Just use a the solver on your calculator .

[DQW]

Bobby, you have an error in your first equation (and several errors subsequenntly).

 

It should read [imath]V = (1/3) \pi r^2 h [/imath], not [imath]V = (1/3) (2 \pi r) h [/imath]

 

Please recheck your equations.

[Pat Says]

One of these is the maximum... the other is the minimum.

Just create a sign chart. Test a couple of points and drop your extraneous points too.

 

It should read V = (1/3) \\pi r^2 h, not V = (1/3) (2 \\pi r) h

Hmm, that's weird because his equation (with the errors) had a max at theta = pi... which is, I'm guessing, the actual answer. Strange coincidence or maybe it just works out the same only with different constants?

[DQW]

There are many more mistakes in his calculation. For one thing (if I'm reading the notation correctly), I don't see any square roots. The heights (calculated from Pythagoras) must involve square roots.

[BobbyJoeCool]

I'll try this again then... (I had 15 minutes before I had to be at work and didn't have time to proofread.)

 

again, we're saying that r=1

 

V=1/3*pi*r'^2*h

 

drawing up the first cone again, the radius of the cone is (2*pi-Ø)/(2*pi) because 2pi-Ø is the circumfrance of the circle at the base, and you divide by 2pi to get back to the radius) Just to simplify things, I'm going to subsitute 2*pi as "x"... r'=(x-Ø)/x

 

V1=1/3*pi*((x-Ø)/x)^2*h

 

h is a leg of a right triangle where c=1, and a=(x-Ø)/x and b=h

 

a^2+b^2=c^2

b^2=c^2-a^2

b^2=1-(x-Ø/x)^2

so h^2=1-(x-Ø/x)^2

and h=sqrt(1-(x-Ø)/x)^2)

 

so V1=1/3*pi*((x-Ø)/x)^2*sqrt(1-(x-Ø)/x)^2)

 

onto the second cone...

 

The second cone has a curcumfrance at the base of Ø, so r'=Ø/x

 

V2=1/3*pi*(Ø/x)^2*h

 

again, h is in a right triangle with a=Ø/x, c=1, and b=h

 

a^2+b^2=c^2

b^2=c^2-a^2

b^2=1-(Ø/x)^2

so h^2=1-(Ø/x)^2

and h=sqrt(1-(Ø)/x)^2)

 

and so

 

V2=1/3*pi*(Ø/x)^2*sqrt(1-(Ø)/x)^2)

 

and so,

 

V1+V2=1/3*pi*((x-Ø)/x)^2*sqrt(1-(x-Ø)/x)^2)+1/3*pi*(Ø/x)^2*sqrt(1-(Ø)/x)^2)

 

or if you like

 

V1+V2=1/3*(pi*((x-Ø)/x)^2*sqrt(1-(x-Ø)/x)^2)+pi*(Ø/x)^2*sqrt(1-(Ø)/x)^2))

 

taking the derivative by hand is still a pain. but this equation still looks wrong, because I don't remember having square roots. I remember when we did this with one of the cones, I had the luxury of squaring both sides of the equation (thus getting rid of the square root), but I'm not sure if it applies here... unless you square the volume of EACH cone before adding them, but would you come up with the same result? I know it's just pi, but it was getting there that was challenging.

[BobbyJoeCool]

how do you guys get that fancy math type that can actually show pi?

[ydoaPs]

[math]\LaTeX[/math] tags. [ math ]...[ /math ] without the spaces

[BobbyJoeCool]

I suppose I could ask what you put in there to get it (as in do you just type pi?) or where can I see something that will show me how to do it, because I couldn't find it in the FAQ...

[ydoaPs]

pi([math]\pi[/math] or [math]\Pi[/math]) is \pi or \Pi, but if you want it for multiplication it is \prod [math]\prod[/math]. there should be a [math]\LaTeX[/math] tutorial stickyd somewhere

[DQW]

Just click on the text and a window will pop up with the script in it.

 

I get (leaving out constants like 1/3, pi, etc) :

 

[math] V' = (1-f)^2 \sqrt{2f - f^2} + f^2 \sqrt{1-f^2} [/math]

 

where [imath] f = \phi /2 \pi [/imath]