Karl's Calculus Tutor - Solution to the Cone Inscribed in the Sphere Problem

Solution to the Cone Inscribed in the Sphere Problem


© 2001 by Karl Hahn
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The problem is choose the dimensions of a cone, height h, and base radius r, in such a way that in can be inscribed in a sphere of radius, R, and maximize the volume of the cone. You have to use some trigonometry to solve this one. The figure on the right shows a cross section of the cone inscribed in the sphere. The point, O, is the center of the sphere. The point, A, is the apex of the cone. The green isosceles triangle is the outline of where the cone intersects the midplane of the sphere. You also have the line segment

   __
   AC  =  

axis of the cone
and

   __
   OD    

bisects angle


  AOB
hence (since triangle AOB is isosceles)

   __
   OD    

intersects

  __
  AB  

at right angles
The angle, θ, is the half-angle of the cone's apex. As is shown in the diagram, you can use the "cosine equals adjacent over hypotenuse" rule to establish that the length


  __
  AD  =  R cos(θ)
Since triangle, AOB, is isosceles, you also get length

  __
  AB  =  2R cos(θ)
which is the hypotenuse of the right triangle, ABC. Using the "sine equals opposite over hypotenuse" rule, you can establish the length

  __
  BC  =  r  =  2R cos(θ)sin(θ)
If you check the trig identity page, you find that  2 cos(θ)sin(θ) = sin(2θ)  (equation 7.1b-6b).

  __
  BC  =  r  =  R sin(2θ)
Likewise, using the "cosine equals adjacent over hypotenuse" rule, you can establish

  __
  AC  =  h  =  2R cos2(θ)
Also from the trig identity page (equation 7.1b-6a), you can easily get  2 cos2(θ) = 1 + cos(2θ).  So:

  __
  AC  =  h  =  R (1 + cos(2θ))

The formula for the volume of a cone is

         1
   V  =    πr2h
         3
Substituting from the above for r and h, this becomes
         1
   V  =    πR3 sin2(2θ)(1 + cos(2θ))
         3
In most min-max problems where you have two or more independent variables (here we have independent variables r and h) you try to solve for one in terms of the other. But here you can begin to see another alternative. We have solved for both r and h in terms of a third, related variable, θ. And we have expressed the dependent variable, volume, V, in terms of θ alone. So now we can find the value of θ that maximizes V. To do the problem that way, you would find the derivative, dV/dθ, and set it to zero.

But taking the derivative of the product of those trig functions is messy. There is an even easier way. Notice from the trig identity,  sin2(2θ) = 1 - cos2(2θ)  the volume equation becomes

         1
   V  =    πR3 (1 - cos2(2θ)) (1 + cos(2θ))
         3
We can get rid of all the trig functions from this simply by substituting a new variable,  u = cos(2θ)

         1
   V  =    πR3 (1 - u2) (1 + u)  =
         3
  1
    πR3 (1 + u - u2 - u3)
  3
Now we take the derivative,  dV/du,  and find the value of u that makes it zero.

   dV           1
       =  0  =    πR3 (1 - 2u - 3u2)
   du           3
Solving the quadratic for u using the quadratic formula:

              ______
         2 ± √4 + 12
   u  =               =  -1
             -6
or

1
 
3
So what do these solutions do for us? Well remember that  u = cos(2θ),  and

   h  =  R (1 + cos(2θ))  =  R(1 + u)
So either


   h  =  0

or
      4
h  =    R
      3
When  h = 0  you get a very uninteresting cone whose volume is zero. That solution certainly does not maximize the volume. So the second solution must be the one that maximizes volume. And I will let you complete the problem by finding

          ______
   r  =  √1 - u2 R


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