The maximum volume of the right circular cone having slant height 3m is:
(A) 3√3π
(B) 6π
(C) 2√3π
(D) (4/3)π

To find the maximum volume of a right circular cone with a given slant height, we need to use the formula for the volume of the cone and apply the condition for maximum volume.

### Step 1: Formula for the volume of a cone
The volume \(V\) of a right circular cone is given by the formula:
V = (1/3)πr²h
where:
- r is the radius of the base of the cone,
- h is the height of the cone,
- π is the constant pi (approximately 3.1416).

### Step 2: Relationship between the radius, height, and slant height
The slant height \(l\) of the cone is related to the radius \(r\) and height \(h\) by the Pythagorean theorem:
l² = r² + h²
We are given that the slant height \(l = 3\) m, so:
3² = r² + h²
9 = r² + h²
This gives us the equation:
h² = 9 - r²
h = √(9 - r²)

### Step 3: Substituting for height in the volume formula
Now, substitute the expression for \(h\) into the formula for the volume:
V = (1/3)πr²√(9 - r²)

### Step 4: Maximizing the volume
To find the value of \(r\) that maximizes the volume, we differentiate the volume function with respect to \(r\) and set the derivative equal to zero.

Differentiate \(V(r) = (1/3)πr²√(9 - r²)\) using the product rule:

dV/dr = (1/3)π [ 2r√(9 - r²) + r² * (-r / √(9 - r²)) ]

Simplify the expression:
dV/dr = (1/3)π [ 2r√(9 - r²) - r³ / √(9 - r²) ]

Set the derivative equal to zero:
2r√(9 - r²) = r³ / √(9 - r²)

Multiply both sides by √(9 - r²) to eliminate the square root:
2r(9 - r²) = r³

Simplify the equation:
2r(9 - r²) = r³
18r - 2r³ = r³
18r = 3r³
6 = r²
r = √6

### Step 5: Calculating the height
Now substitute r = √6 into the equation \(h = √(9 - r²)\):
h = √(9 - 6)
h = √3

### Step 6: Finding the maximum volume
Substitute r = √6 and h = √3 into the volume formula:
V = (1/3)πr²h
V = (1/3)π(6)(√3)
V = 2√3π

### Step 7: Conclusion
The maximum volume of the cone is 2√3π cubic meters.

The correct answer is: **C) 2√3π**.