MathematicsThe radius of a sphere is  9cm  . Find the height of the right circular cone that can be inscribed in the sphere so that the volume of the cone is maximum.

The radius of a sphere is  9cm  . Find the height of the right circular cone that can be inscribed in the sphere so that the volume of the cone is maximum.


  1. A
    h=12cm
  2. B
     h=15cm
  3. C
    h=10cm
  4. D
    h=11cm 

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    Solution:

    Let us first draw the diagram.
    IMG_256Let the height of the cone be  h   and the radius of the cone be r  . From the diagram, we can see that  OAB   forms a right-angled triangle. Since  OB   represents the radius of the sphere, therefore OB=9cm  .
    Also,  OC   represents the radius of the sphere, therefore  OC=9cm  .
    As we can see that  AC=h   and  AC=AO+OC  , therefore
    h=AO+9 AO=h9  
    In right angles triangle ΔOAB,AO=h9,BO=9cm  .
    Therefore, using Pythagoras’ Theorem,
    BO 2 = AO 2 + AB 2 9 2 = h9 2 + AB 2 AB 2 =81 h9 2  
    AB 2 =81 h 2 81+18h AB 2 =18h h 2  
    As we can see ABAB represents the radius of the base of the cone, therefore  r 2 =18h h 2  .
    Now the volume of the cone is given by ,
    V= 1 3 π r 2 h  
    where  r   is the radius of the base of the cone and  h   is the height of the cone.
    Therefore,
    V= 1 3 π 18h h 2 h V= 1 3 π 18 h 2 h 3   Let us take the derivative of  V   with respect to  h  . We will get,
    dV dh = 1 3 π 36h3 h 2  
    Taking dV dh =0   will give us two values of  h  one of them gives the maximum volume of the cone and the other one gives the minimum volume of the cone.
    Therefore,
    1 3 πh 363h =0  gives h=0   and 363h  that is h=12cm  .
    Taking derivative of  dV dh  , we get
    d 2 V d h 2 = 1 3 π 366h  
    When we put  h=12cm  in  d 2 V d h 2  , we get
    d 2 V d h 2 h=12cm = 1 3 π 366(12) d 2 V d h 2 h=12cm = 1 3 π36  
    d 2 V d h 2 h=12cm =12π  ,which is negative.
    Therefore,  h=12cm   represents the maximum volume of cones inscribed in the sphere.
     
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