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Drinking cup

  1. Apr 20, 2007 #1
    1. The problem statement, all variables and given/known data
    A cone-shape drinking cup is made from a circular piece of paper of radius R by cutting out a sector and joining the edges CA and CB. Find the maximum capacityof such a cup. This is page#312 In James stewarts Calc book, 3rd edition by the way.

    2. Relevant equations
    [tex]V=\frac{1}{3}\pi r^2 h[/tex]

    3. The attempt at a solution
    So realize this will form a cone with an inner triangle of hypotnue R. The base will have a radius r and a hieght h. So I will have to make a relationship between r and h to get R to plug into [tex]V=\frac{1}{3}\pi h r^2[/tex] and differeniate?
     
    Last edited: Apr 20, 2007
  2. jcsd
  3. Apr 20, 2007 #2

    Dick

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    (1/3)*pi*r^2 doesn't even have the units of a volume. (1/3)*pi*r^2*h does. I think that's what you want. You will want to parametrize r and h by the angle cut out of R. And then differentiate and maximize wrt to the angle.
     
  4. Apr 20, 2007 #3
    sorry clumsy mistake, I forgot the H, lol.
     
  5. Apr 20, 2007 #4

    Dick

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    I thought so. But just write r and h as functions of the cut out angle. That's just geometry.
     
  6. Apr 21, 2007 #5

    HallsofIvy

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    Notice that the entire circle of paper has circumference [itex]2\pi R[/itex] but the circle at the top of the cone has circumference [itex]2\pi r[/itex]. The arc length of the cut out wedge is [itex]R \theta/\2\pi[/itex] where [itex]\theta[/itex] is the angle in radians. That must be the difference between [itex]2\pi R[/itex] and [itex]2\pi r[/itex].
     
  7. Apr 21, 2007 #6
    mmmm.......
     
  8. Apr 22, 2007 #7

    Dick

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    Ok, then let's try it another way. Draw the right triangle joining a point on the circle of the base of the cone to the apex of the cone. Height is h and base is r, right? What's the length of the hypotenuse? Use Pythagoras to eliminate one of the variables (r or h) from the formula for the volume and maximize.
     
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