The height of the cylinder of maximum volume which can be inscribed in a sphere of radius `3 cm` is

Options

A. `sqrt(3)`
B. `2sqrt(3)`
C. `(2sqrt(3))/(3)`
D. `3sqrt(2)`

Answer

To find the height of the cylinder of maximum volume that can be inscribed in a sphere of radius \( R = 3 \, \text{cm} \), we can follow these steps:

### Step 1: Understand the Geometry
We have a sphere of radius \( R \) and a cylinder inscribed within it. Let the radius of the cylinder be \( r \) and the height of the cylinder be \( h \). The center of the sphere is at point \( O \), and the top and bottom of the cylinder touch the sphere.

### Step 2: Relate the Cylinder and Sphere Dimensions
Using the Pythagorean theorem on the triangle formed by the center of the sphere, the radius of the cylinder, and half the height of the cylinder:
\[ R^2 = r^2 + \left(\frac{h}{2}\right)^2 \]
Substituting \( R = 3 \):
\[ 3^2 = r^2 + \left(\frac{h}{2}\right)^2 \]
This simplifies to:
\[ 9 = r^2 + \frac{h^2}{4} \]

### Step 3: Express Volume of the Cylinder
The volume \( V \) of the cylinder is given by:
\[ V = \pi r^2 h \]
Using the equation from Step 2, we can express \( r^2 \) in terms of \( h \):
\[ r^2 = 9 - \frac{h^2}{4} \]
Substituting this into the volume formula:
\[ V = \pi \left(9 - \frac{h^2}{4}\right) h \]
Thus,
\[ V = \pi \left(9h - \frac{h^3}{4}\right) \]

### Step 4: Differentiate Volume with Respect to Height
To find the maximum volume, we differentiate \( V \) with respect to \( h \):
\[ \frac{dV}{dh} = \pi \left(9 - \frac{3h^2}{4}\right) \]
Setting the derivative equal to zero for maximization:
\[ 9 - \frac{3h^2}{4} = 0 \]

### Step 5: Solve for Height \( h \)
Rearranging gives:
\[ \frac{3h^2}{4} = 9 \]
Multiplying both sides by \( \frac{4}{3} \):
\[ h^2 = 12 \]
Taking the square root:
\[ h = 2\sqrt{3} \, \text{cm} \]

### Conclusion
Thus, the height of the cylinder of maximum volume that can be inscribed in a sphere of radius \( 3 \, \text{cm} \) is:
\[ \boxed{2\sqrt{3} \, \text{cm}} \]