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Homework Help: Find largest possible volume (Extreme Value)

  1. Feb 20, 2016 #1
    • Member warned about posting homework in non-homework forum sections
    Mod note: Moved from technical math section, so is missing the homework template.
    A right triangle the hypotenuse of which is c revolves about a small side. A right circular cone is formed. Find the largest possible volume of the cone.

    Vcone: pi*r^2*(h/3)

    I don't get why they mentioned the hypotenuse, since it is not even in the volume formula...
    Since it is a 'right' triangle, shouldn't the maximum height of the cone be 1?
    But I still don't know which extreme value I should search for...
    (The solution should be: 1/(9*sqrt3) * c^3 * pi )

    Could someone please help me to?
    Last edited by a moderator: Feb 20, 2016
  2. jcsd
  3. Feb 20, 2016 #2


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    You mentioned the solution which obviously has a ##c## in it. So probably the hypotenuse will play a role.
    How do you derive that ##h=1##?
    As I see it from your solution you are supposed to maximize ##V_{cone}## for a given length of the hypotenuse ##c##.
    Clearly ##c=0## is a minimum.
    How do you usually calculate maxima?
    Last edited: Feb 20, 2016
  4. Feb 21, 2016 #3
    We only learned to make the derivative equal to zero in order to find it.
  5. Feb 21, 2016 #4
    I think there is a mistake in the answer which you will find if you work through it.

    If you know that the sloping side of the cone is the hypotenuse of a right angled triangle length c, can you find a relationship between c, r and h?
  6. Feb 21, 2016 #5
    isn't it
    c^2= r^2+h^2 ?
    r^2 = c^2 - h^2

    Then I could substitute it:

    V = pi * (c^2-h^2) * h/3
    V = pi * ( (c^2h/3) - (h^3/3) )
    V = pi * h/3 * (c^2 - h^2)


    pi * h/3 * d/dc (c^2 - h^2)
    = 2/3 * pi * h * c

    But this isn't the solution... What did I do wrong?
  7. Feb 21, 2016 #6


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    The fact that they say "the hypotenuse of which is c" tells you that the hypotenuse is a constant. You certainly should NOT be differentiating with respect to c!
  8. Feb 21, 2016 #7
    You're doing fine up to here, let me write it more clearly as $$ V = \frac{\pi c^2}3 h - \frac 13 h^3 $$So taking on board HallsOfIvy's comment, what's next?
  9. Feb 22, 2016 #8

    Ray Vickson

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    No. The equation
    [tex] V = \frac{\pi c^2}3 h - \frac 13 h^3 [/tex]
    is not correct. However, the equation
    [tex] V = \frac{\pi}{3} \left( c^2 h - h^3 \right) [/tex]
    is correct.
  10. Feb 22, 2016 #9
    Ah thanks for fixing that, I wanted to split up the terms to make the differentiation more obvious but made a mess of it!
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