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Homework Help: Finding the maximum volume of a cone

  1. Oct 26, 2009 #1
    1. The problem statement, all variables and given/known data

    In England, you can purchase fish and chips for a reasonable price. The reason it is so reasonable is because they give you no silverware, nor a plate. They just roll up a piece of paper in a cone and toss your food in. The vendors need to roll the cone in a perfect size, not too fat nor too skinny. Find out how to optimize the volume of the cone. For modeling purposes, assume that the piece of paper is a circle of radius 5 inches, and that we are cutting a wedge out of it whose central angle is Θ. Find the maximum volume of this cone.

    2. Relevant equations

    A = 2πr

    i need to find out a formula to subtract the cut out wedge of the circle from the rest of the circle.

    3. The attempt at a solution

    I am stuck trying to find a main formula. I would think that after i get that formula, i find the derivative and then set it equal to zero, and solve.
  2. jcsd
  3. Oct 26, 2009 #2


    Staff: Mentor

    Your relevant equation is irrelevant. That's not the area of a circle; it's the circumference. Some actually relevant equations that you should look up are the area of a circle, the volume of a cone, and the area of a sector of a circle. See if you can find those in your textbook.
    Last edited: Oct 26, 2009
  4. Oct 27, 2009 #3
    ok, i found out the volume forumula
    i divided the cone into 2 right triangles, using 5 as the hypotenuse, and r as the radius of the base of the cone.

    so i have

    V = (1/3)(pi)(square root of 25 - x squared)(x squared)

    i am working on finding the derivative but there is so much work with the product rule, chain rule, and quotient rule, i think i may have screwed up somewhere

    so far i am at,

    dV/dx = [(1/3)(pi)][(1/2)(25-x^2)^(-1/2) *-2x] * 2x + x^2 * (derivative of first part)
    Last edited: Oct 27, 2009
  5. Oct 27, 2009 #4


    Staff: Mentor

    This really doesn't make any sense. A cone is a 3D object. How can you divide it up into two triangles, which are 2D objects.
    You're jumping into this problem without setting yourself up with the formulas you need. The volume of a right circular cone (what you're dealing with) is V = 1/3 * pi * r^2 * h. That's one of your relevant formulas.

    Another one is the area of a sector of a circle, sort of like a wedge of pie of any size. This formula is A = (1/2)[itex]\theta r^2[/itex], where [itex]\theta[/itex] is measured in radians.
  6. Oct 27, 2009 #5
    V = 1/3 * pi * r^2 * h

    thats what i had, but i substituted in h in terms of x but using pythagorean.

    x^2 + h^2 = 5 ^2

    h^2 = 25 - x^2
    h = square root of (25 - x^2)

    http://z.about.com/d/math/1/5/t/L/conerr.jpg [Broken]
    Last edited by a moderator: May 4, 2017
  7. Oct 27, 2009 #6


    Staff: Mentor

    Even with the picture, this formula still doesn't make any sense. You used 5 for the radius. What does x represent? What does sqrt(25 - x^2) represent?
    You're also not taking into account that the circular piece of paper has a wedge cut out of it.
  8. Oct 27, 2009 #7
    the sqrt(25 - x^2) represented h. x represented the radius (r in the picture). 5 represented the S on the picture.

    i thought the A = (1/2)theta r^2 would be irrelevent because i'm trying to find volume, not area?
  9. Oct 27, 2009 #8


    Staff: Mentor

    OK, now it makes sense. It would have helped me understand better if you had explained what your were doing to get the formula you showed.

    The 5 dimension in the drawing is what is called the slant height of the cone. Here's your formula again, this time with r instead of x. r matches your diagram and suggests what it represents, the radius of the cone.

    [tex]V(r)~=~(1/3)\pi r^2(25 - r^2)^{1/2}[/tex]

    Now you want to take the derivative. You'll need to use the product rule and then the chain rule, in that order.
  10. Oct 27, 2009 #9
    ok cool.

    so v'(r) = (1/3)pi(r^2)*(-1/(2*sqrt(25 - x^2)) + [(25-r^2)^(1/2)]*[(2/3)*pi*r]

  11. Oct 27, 2009 #10


    Staff: Mentor

    No. In the chain rule, you forgot the factor that comes from d/dr(25 - r^2). Also, since V is now a function of r, then V'(r) shouldn't have any x's floating around in it.
  12. Jan 20, 2011 #11
    Ok so if you don't want to use calculus you can AM-GM it.
    So the volume of the cone would be pi(r^2)h/3.
    Arc length=circumference of cone=theta/360*10pi
    so the radius of the cone=(theta/360*10pi)/2pi = theta/72
    the slant height of the cone is 5 so the height is sqrt(25-(theta/72)^2)
    so the volume of the cone is pi(theta/72)^2*sqrt(25-(theta/72)^2)/3
    now all we have to do is find the maximum of this equation
    we can get rid of the /3 and pi because those are constant
    if we let x=(theta/72)^2 then the formula becomes (x)sqrt(25-x)
    much simpler, eh?
    this formula is maximized when (x^2)(25-x) is maximized (for obvious reasons)
    so now we can use AM-GM on the formula (x^2)(50-2x), which is maximized when the three terms are closest to eachother
    so we let x=x=50-2x and 50/3
    going back to what we defined x to be, x=50/3=(theta/72)^2
    theta=293.939 degrees and you're done!
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