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Maximize volume of a cone, how?

  1. Mar 10, 2005 #1
    Hello, I have a problem I cant solve. Need assistance! :bugeye:

    You cut out a piece of a circle (like you cut a piece of a cake), then make a cone by joining the edges of what remains of the circle. What angle must the "cakepiece" have to maximize the volume of the resulting cone?

    I know the cone Volume is V = pi*r^2h/3, and I need to derive this to get maxpoints. Cant figure out how though?

    question 2.

    this time, we also make a cone out of the piece we cut out. What angle must this piece have to maximize the volume of BOTH the cones made?

    Please point me in the right direction if you will. Thanx a lot! :redface:
  2. jcsd
  3. Mar 10, 2005 #2


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    1.Terminology.It's a disk,not a circle...A circle is a curve...
    2.The second problem is really interesting.To the first i've got the angles for extremum
    [tex] \alpha_{1}=0;\alpha_{2}=+\pi\sqrt{\frac{8}{3}};\alpha_{3}=-\pi\sqrt{\frac{8}{3}} [/tex]

    Sine the 0 is not okay,you're left with 2.The positive is okay.It's a bit greater than \pi...The negative solution would correspond physically to an angle of [tex] 2\pi-\pi\sqrt{\frac{8}{3}} [/tex],but the volume found by plugging in its expression would be certainly less...


    HINT:Denote by "r" the radius of the disk from which u cut.That will be the generator of the cone...This "r" is assumed constant.Denote by "a" the length of the arch subtended on the disk to the angle [itex] \alpha [/itex].In the same manner,denote by "h" the height of the cone and by R the radius of the basis circle.

    Work out all possible relations between these variables (they're only 3) and then plug in the formula

    [tex] V=\frac{\pi R^{2}h}{3} [/tex] everything in terms of the constant "r" and the angle "alpha"...

    Then set the first derivative to 0,to find the 3 solutions i posted above.
  4. Mar 10, 2005 #3
    thanks for the help mate, appreciate it. Im not native in english, but of course disc was the word to use :redface:

    Im still having problems figuring this out.

    the section cut away is [tex]\alpha*r[/tex]
    the remaining circumference of the disc, thus the base of the cone is
    [tex](a - \alpha*r)[/tex]

    The side of the cone is [tex]r[/tex]
    the height is h
    the radius of the cone is [tex]R= (a - \alpha*r)/2\pi[/tex]

    so V = [tex]((a - \alpha*r)/2\pi))^2 * h\pi/3 [/tex] ?

    I have too many unknowns? I know theres an easy answer but I just can't see it. It's late =D

    Care to give me one more hint?

    again thanks for helping out
  5. Mar 10, 2005 #4


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    Sorry.I missunderstood the problem.I took into account only what's cut of the disk.I mean from the piece that u remove you build the cone.

    I'll think about it and redo the calculations.

  6. Mar 10, 2005 #5


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    Use Pythagora's theorem for the right trangle:

    [tex] R^{2}+h^{2}=r^{2} [/tex]

  7. Mar 10, 2005 #6
    But that still leaves too many unknowns?

    I need both the radius of the cone. the height. and the angle (which is what im looking for really).

    I will continue in the morning, doing this at midnight is not a good idea =).
    thanx for helping
  8. Mar 10, 2005 #7


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    Nope,it leaves the angle & the radius.Since the radius of the initial disk is constant,the volume function will depend on one variable only:the angle of the center of the disk...

  9. Mar 11, 2005 #8
    Ok ive been working with this. I thought I would make the algebra easier by deciding the radius of the disc to be 1. Since it's constant it shouldnt matter which value i put here right.

    That gave me the following:
    R = 1

    [tex]V = \pi/3(1-X/2\pi)^2 * (1-(1-X/2\pi))[/tex]

    [tex]V' = 2\pi/3 * (1-X/2\pi) * (-1/2\pi) * (1-(1-X/2\pi)) + \pi/3(1-X/2\pi)^2 * (-1/2\pi)


    Now I think I can get this equation to 0 by setting X to be [tex]-2\pi[/tex]
    That gave me a value of X = 951,... which cant be right. where am I going the wrong way? I just can't see it. Are my calculations correct so far?

    Last edited: Mar 11, 2005
  10. Mar 11, 2005 #9
    For starters, let the radius of the disk be R. Then the resulting cone will be a volume of revolution of a right triangle whose hypoteneuse is R. Intuition should tell you that the triangle with the largest area gives you the cone with the largest volume. I will leave out the proof of this, but there is a formula that relates the area of a shape to the volume of revolution for that shape which should do the trick. Intuition (and symmetry) should convince you that the largest area for such triangles is achieved when the two remaining sides are equal. I will leave out this proof as well. However, I will give an outline. Call the length of the vertical side h, and the length of the horizontal side r. Then

    [tex]h = \sqrt{R^2 - r^2}[/tex]

    The area of the triangle is:

    [tex]A = (1/2)rh = (1/2)r\sqrt{R^2 - r^2}[/tex]

    Take the derivative of A and set it to zero. This will give

    [tex]r = R\sqrt{1/2}[/tex]


    [tex]h = R\sqrt{1/2}[/tex]

    so that r = h as expected.

    The circumference of the disk is [tex]2\pi R[/tex]. When you take out the slice of angle [tex]\alpha[/tex], the length of the remaining part of the circumference will be

    [tex]R(2\pi - \alpha)[/tex]

    When the disk is deformed to make the cone, this curve will be bent into a circle. The radius of the circle will be the length of the curve divided by [tex]2\pi[/tex]. That is:

    [tex]r = R(2\pi - \alpha)/(2\pi)[/tex]

    Since we want [tex]r = R\sqrt{1/2}[/tex] we set:

    [tex]R(2\pi - \alpha)/(2\pi) = R\sqrt{1/2}[/tex]

    R drops out (as expected) and we get

    [tex]\alpha = \pi(2 - \sqrt{2})[/tex]

    I didn't do the second part, but intuition and symmetry tell me to cut the disk in half. I have no proof for this.
    Last edited: Mar 11, 2005
  11. Mar 12, 2005 #10
    The optimum angle is 1.84... radians then? Will try your method to see if i fully understand it. I'm writing a small paper on this so I might need to find the proof that max volume is given by that triangle. I can find that myself though probably. Thanx. Then i still have question 2 to figure out =D
  12. Mar 12, 2005 #11
    In your paper you might want to mention the fact that when you deform the disk with the slice taken out, you do not get a right cone. There might be some contraversy concerning what is and what is not included in the volume of the object that you do get. Therefore, the method of proof and the result should be taken as approximations.
    Last edited: Mar 12, 2005
  13. Mar 12, 2005 #12

    I think Ive got it. Could you please explain how you derived the above. I dont fully understand how to derive when there are square roots involved.

  14. Mar 12, 2005 #13
    I have the answer now and it seems 1.84 radians was correct!

    Question number two is next. Is it possible to set V = V1 + V2 and just derive that do you think?
  15. Mar 13, 2005 #14
    Before I derive the give formula, I will show how to derive when there are square roots involved. The trick is to express the square root as a fractional power:

    [tex]\sqrt{x} = x^\frac{1}{2}[/tex]

    Then use the well known formula for [tex]f(x) = x^n[/tex]

    [tex]f'(x) = nx^{n-1}[/tex]

    so since [tex]f(x) = x^\frac{1}{2}[/tex]

    [tex]f'(x) = \frac{1}{2}x^\frac{-1}{2}[/tex]

    Now for the derivation

    [tex]A = \frac{1}{2}r\sqrt{R^2 - r^2} = \frac{1}{2}r(R^2 - r^2)^\frac{1}{2}[/tex]

    [tex]A' = \frac{1}{2}r\frac{1}{2}(R^2 - r^2)^\frac{-1}{2}(-2r)
    + \frac{1}{2}(R^2 - r^2)^\frac{1}{2}
    = \frac{1}{2}(R^2 - r^2)^\frac{-1}{2}(-r^2 + R^2 - r^2)[/tex]

    setting this to zero we get:

    [tex]0 = R^2 - 2r^2[/tex]

    or [tex]r = \frac{1}{\sqrt{2}}R[/tex]
    Last edited: Mar 13, 2005
  16. Nov 8, 2011 #15
    This is incorrect.

    Here is a correct solution:

    http://mathcentral.uregina.ca/QQ/database/QQ.09.06/s/christina1.html" [Broken]
    Last edited by a moderator: May 5, 2017
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