The maximum volume (in cu.m) of the right circular cone having slant height 3 m is

Options

A. `3sqrt3 pi`
B. `6pi`
C. `2sqrt3pi`
D. `4/3pi`

Answer

To find the maximum volume of a right circular cone with a slant height of 3 m, we will follow these steps:

### Step 1: Understand the relationship between radius (R), height (H), and slant height (L)

The relationship is given by the Pythagorean theorem:
\[ L^2 = R^2 + H^2 \]
Given that the slant height \( L = 3 \) m, we can write:
\[ 3^2 = R^2 + H^2 \]
This simplifies to:
\[ R^2 + H^2 = 9 \]

### Step 2: Write the volume formula for the cone

The volume \( V \) of a right circular cone is given by:
\[ V = \frac{1}{3} \pi R^2 H \]

### Step 3: Substitute \( R^2 \) in terms of \( H \)

From the equation \( R^2 + H^2 = 9 \), we can express \( R^2 \) as:
\[ R^2 = 9 - H^2 \]
Now substitute this into the volume formula:
\[ V = \frac{1}{3} \pi (9 - H^2) H \]
This simplifies to:
\[ V = \frac{\pi}{3} (9H - H^3) \]

### Step 4: Differentiate the volume with respect to height \( H \)

To find the maximum volume, we need to differentiate \( V \) with respect to \( H \) and set the derivative equal to zero:
\[ \frac{dV}{dH} = \frac{\pi}{3} (9 - 3H^2) \]
Setting this equal to zero gives:
\[ 9 - 3H^2 = 0 \]

### Step 5: Solve for \( H \)

Rearranging the equation:
\[ 3H^2 = 9 \implies H^2 = 3 \implies H = \sqrt{3} \]
Since height cannot be negative, we take \( H = \sqrt{3} \) m.

### Step 6: Substitute \( H \) back to find \( R \)

Now, substitute \( H \) back into the equation for \( R^2 \):
\[ R^2 = 9 - H^2 = 9 - 3 = 6 \implies R = \sqrt{6} \]

### Step 7: Calculate the maximum volume

Substituting \( H \) and \( R \) back into the volume formula:
\[ V = \frac{1}{3} \pi R^2 H = \frac{1}{3} \pi (6)(\sqrt{3}) = \frac{6\sqrt{3}}{3} \pi = 2\sqrt{3} \pi \]

Thus, the maximum volume of the cone is:
\[ \boxed{2\sqrt{3} \pi} \text{ cubic meters} \]