Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is 4r/3. Also, find maximum volume in terms of volume of the sphere.

Answer

To solve the problem of finding the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius \( r \), and to express the maximum volume in terms of the volume of the sphere, we can follow these steps:

### Step 1: Understand the Geometry
Consider a sphere of radius \( r \) and a right circular cone inscribed within it. Let the radius of the base of the cone be \( R \) and the height of the cone be \( h \).

### Step 2: Relate the Cone's Dimensions to the Sphere
Using the Pythagorean theorem in the triangle formed by the radius of the sphere, the radius of the cone, and the height of the cone, we have:
\[ r^2 = R^2 + (r - h)^2 \]
Expanding this, we get:
\[ r^2 = R^2 + r^2 - 2rh + h^2 \]
This simplifies to:
\[ 0 = R^2 - 2rh + h^2 \]
Rearranging gives:
\[ R^2 = 2rh - h^2 \]

### Step 3: Volume of the Cone
The volume \( V \) of the cone is given by:
\[ V = \frac{1}{3} \pi R^2 h \]
Substituting for \( R^2 \) from the previous step:
\[ V = \frac{1}{3} \pi (2rh - h^2) h \]
This simplifies to:
\[ V = \frac{1}{3} \pi (2rh^2 - h^3) \]

### Step 4: Differentiate the Volume
To find the maximum volume, we differentiate \( V \) with respect to \( h \):
\[ \frac{dV}{dh} = \frac{1}{3} \pi (4rh - 3h^2) \]
Setting the derivative equal to zero to find critical points:
\[ 4rh - 3h^2 = 0 \]
Factoring gives:
\[ h(4r - 3h) = 0 \]
Thus, \( h = 0 \) or \( h = \frac{4r}{3} \).

### Step 5: Verify Maximum Volume
To confirm that \( h = \frac{4r}{3} \) gives a maximum, we can check the second derivative:
\[ \frac{d^2V}{dh^2} = \frac{1}{3} \pi (4r - 6h) \]
Evaluating at \( h = \frac{4r}{3} \):
\[ \frac{d^2V}{dh^2} = \frac{1}{3} \pi (4r - 6 \cdot \frac{4r}{3}) = \frac{1}{3} \pi (4r - 8r) = \frac{1}{3} \pi (-4r) \]
Since this is negative, \( h = \frac{4r}{3} \) is indeed a maximum.

### Step 6: Calculate Maximum Volume
Substituting \( h = \frac{4r}{3} \) back into the volume formula:
\[ R^2 = 2r \left(\frac{4r}{3}\right) - \left(\frac{4r}{3}\right)^2 = \frac{8r^2}{3} - \frac{16r^2}{9} = \frac{24r^2 - 16r^2}{9} = \frac{8r^2}{9} \]
Thus, the volume becomes:
\[ V = \frac{1}{3} \pi \left(\frac{8r^2}{9}\right) \left(\frac{4r}{3}\right) = \frac{32 \pi r^3}{81} \]

### Step 7: Express in Terms of Sphere Volume
The volume of the sphere is:
\[ V_{sphere} = \frac{4}{3} \pi r^3 \]
Now, we can express the maximum volume of the cone in terms of the volume of the sphere:
\[ V = \frac{32 \pi r^3}{81} = \frac{32}{81} \cdot \frac{4}{3} \pi r^3 = \frac{32}{81} V_{sphere} \]

### Final Result
The altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius \( r \) is \( \frac{4r}{3} \), and the maximum volume of the cone is \( \frac{32}{81} \) times the volume of the sphere.