Inscribing a Cone within a Sphere (Could someone double check my work)

[Sacredify]

I find myself a little confused as what answer I got for this problem. It would be cool if someone could look over it for me and maybe correct it if it is wrong Thanks.

The problem: Find the max dimensions of a cone inscribed in a sphere of radius 36.

How I did it: Maximum V in terms of h and r to solve for the height first, then use that to get the radius of the cone.

Little picture to help visualize:



So, we know the equation for volume of a cone is V = pi/3r^2h
For this question, let "x" represent the radius of the cone.

Therefore, Vcone = pi/3x^2h

By the Pythagorean theorem, we know that a^2 + b^2 = c^2

If we look at the diagram, we know that the hypotenuse would be the radius of the sphere(36cm). The other two we do not know. We do know, however, that one would be x, the radius of the cone, and the other h-r, the height of the cone minus the radius.

So,
r^2 = x^2 + (h-r)^2
r^2 = x^2 + h^2 -2hr + r^2
x^2 = 2hr - h^2

Back to the volume equation, and sub this in for x^2

V = pi/3(2hr-h^2)h
= pi/3 (2h^2r-h^3) (at this point, I subbed in 36 for the radius of the sphere, which was given)
= pi/3 (72h^2-h^3)

Now for the derivative:

V' = pi/3 (144h-h^2)
= pi/3(3h(48-h)
Therefore, h = 48

Using that, I plugged the number back into the theorem to solve for the radius of the cone:

36^2 = x^2 + (48-36)^2
x^2 = 1152
x = 33.94


So, the dimensions of the cone were radius 33.94, height 48.

However, to me, the radius sounds a bit off. If the max radius of the sphere is 36, and the height is 48, how is the radius 34? Any response would be appreciated.

 

[soroban]

Hello, Sacredify!

Find the dimensions of a cone of maximum volume inscribed in a sphere of radius 36.

                A
              * o *
          *     |     *
        *       |       *
       *        |36      *
                |
      *         |O        *
      *         o         *
      *       * | *  36   *
            *  y|   *
       *  *     |  x  *  *
        o - - - + - - - o
       B  *     D     *  C
              * * *

\(\displaystyle O\) is the center of the sphere.
Draw \(\displaystyle AB\) and \(\displaystyle AC.\)

\(\displaystyle \Delta ABC\) is the side view of the cone.
Its radius is \(\displaystyle x = DC\); its height is: \(\displaystyle AD = 36 + y\)

In right triangle \(\displaystyle ODC\!:\;x^2+y^2 \:=\:36^2 \quad\Rightarrow\quad y \:=\:\sqrt{1296 - x^2}\)
. . Hence, the height is: .\(\displaystyle h \:=\:36 + \sqrt{1296-x^2}\)


The volume of a cone is: .\(\displaystyle V \:=\:\frac{\pi}{3}r^2h\)

We have: .\(\displaystyle V \;=\;\frac{\pi}{3}x^2\left(36 + \sqrt{1296-x^2}\right)\)

And that is the function we must maximize.

 

[Sacredify]

Was my method wrong? I can see where the equation came from, but how is it any different that solving for the height first?

I looked around on how to do it, and most use the method I (tried) to use. It avoids needlessly complicated derivatives and such.


According to multiple websites, the final answers for this type of problem are

h = 4/3r
r = 2sqrt(2) * r/3

which are my answers...