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Find the Magnitude of Theta to Maximize Volume of a Cone

  1. Mar 18, 2017 #1
    1. The problem statement, all variables and given/known data
    A sector with central angle θ is cut from a circle of radius R = 6 inches, and the edges of the sector are brought together to form a cone. Find the magnitude of θ such that the volume of the cone is a maximum.

    2. Relevant equations
    Volume of a Cone = ⅓ * π * r2 * h

    Area of a Sector of a Circle = R2/2 * Θ

    3. The attempt at a solution
    So, this problem is a lot like other cone/volume problems I have done before, but I have never had to find the magnitude of Theta before.

    Normally, I would use the Pythagorean Theorem to determine what to sub in for "r" in the Volume of a Cone equation (in this case, that would be 36-h2 = r which can be determined by using the Pythagorean theorem as R2 = r2 + h2. In this problem "R" always = 6).

    I know I can use this information to determine the critical numbers of "h," but I can't seem to determine when the Area of a Sector of a Circle would come into play. It is related to "R," which is always 6 in this problem, so I determined that this equation could be simplified to 18*Θ.

    I attempted a solution by finding the critical numbers as I mentioned above and plugged those into Θ, but that doesn't really make sense to me. Any guidance would be appreciated. I don't want to be given the answer, obviously -- This isn't the forum for that. But, any hint or clue would be amazing.

    Thank you so much.
     
  2. jcsd
  3. Mar 18, 2017 #2

    mfb

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    To find theta with your optimal value of r (or h), think about the circumference of the cone base. Where does it come from? How is the length of this related to the original circle with a section cut out?
     
  4. Mar 18, 2017 #3
    Is r = sin(Θ½)? I had thought of that, but I think I confused myself by solving for 18Θ early on.
     
  5. Mar 18, 2017 #4

    mfb

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    r cannot be the sine of anything, that wouldn't work in terms of dimensions - there is no reason why r should be smaller than 1.
    I have no idea where 18Θ comes from.
     
  6. Mar 18, 2017 #5
    The 18Θ came from the Area of a Sector of a Circle formula (R2*½*Θ).
     
  7. Mar 18, 2017 #6

    mfb

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    That area is related to the cone surface area, which you don't need.
     
  8. Mar 18, 2017 #7

    Ray Vickson

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    Avoid plugging in ##R = 6## until the last minute; that way, you can tell where the various numbers are coming from. Not only that, you get a more general formula.

    Anyway, if you cut out a sector of angle ##\theta## from a circle of radius ##R##, then fold it up into a cone, what is the radius of the base of the cone (call it ##r##)?

    Now you have the cone's base ##r## and slanted-side length ##R##, so you can determine its height ##h## and thus find its volume.
     
    Last edited: Mar 19, 2017
  9. Mar 18, 2017 #8
    Hmm... 2π - ##\theta##?

    So, am I better off approaching this in Pythagorean form using ##r##(2π - ##\theta##); ##R##; and ##h##2 = ##R##2 - ##r##2?
     
  10. Mar 19, 2017 #9

    haruspex

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    That cannot be a radius.

    The circumference of the original full circle is 2πR, right?
    After the sector is removed, what length of that remains?
    When formed into a cone, what happens to that part-circle?
     
  11. Mar 19, 2017 #10
    Sorry, I wrote the correct term further down in my post ##r##(2π - ##\theta##).

    So, if that's correct, I should be working with: ⅓(2π##r## - ##\theta## ##r##)2π√(##R##2 - ##r##2)
     
  12. Mar 19, 2017 #11

    haruspex

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    What is that the answer to?
    Ray defined r as the radius after forming the cone. You want an equation of the form r= some function of R and θ.
     
  13. Mar 19, 2017 #12

    mfb

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    That is the wrong radius.
     
  14. Mar 19, 2017 #13
    Ray wrote: "Anyway, if you cut out a sector of angle θ from a circle of radius R, then fold it up into a cone, what is the radius of the base of the cone (call it r)?"

    The radius of the base of the cone would definitely be ##r##2 = ##R##2 - ##h##2. To make this relate to the angle of ##\theta##, the circumference of the base of the cone would be 2π##r## - ##\theta## ##r##, so am I plugging in √(##R##2 - ##h##2) for ##r##?
     
  15. Mar 19, 2017 #14

    Ray Vickson

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    OK, if you know ##h## you can figure out ##r##. However, who tells you the value of ##h##? All you know is ##\theta##.
     
    Last edited: Mar 19, 2017
  16. Mar 19, 2017 #15
    I don't know ##h## or ##r## (except in the formulaic terms mentioned above), I am looking for ##\theta##, and all I am given is ##R## (the radius of the circle from which the sector is cut).
     
  17. Mar 19, 2017 #16

    Ray Vickson

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    Exactly; but you can work out a formula for ##r## in terms of ##R## and ##\theta##. (You tried that before, but made some errors.)

    In the end, you will get a formula for the volume ##V## in terms of ##R## and ##\theta##, and since ##R## is just a constant input parameter, the only variable involved is ##\theta##.
     
  18. Mar 19, 2017 #17

    haruspex

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    Eh? You have a circle radius r. What is its circumference?
     
  19. Mar 19, 2017 #18

    mfb

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    @vr0nvr0n: You are confusing yourself by trying too many steps at once.

    What is the circumference of the base of the cone - a full circle with radius r?

    Going back to the original circle with radius R with a section cut out: How large was the original circumference, how large is the remaining part of the circumference?
     
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