If a right circular cone, having maximum volume, is inscribed in a sphere of radius 3 cm, then the curved surface area (in cm²) of this cone is

Sphere of radius : 3 cm (r=3)
Let b, h be radius and height of sphere, respectively.
∴ volume of cone = (1/3)πb²h
In ΔABC, using Pythagoras theorem
(h−r)² + b² = r² ……(i)
b² = r² − (h−r)² = r² − (h² − 2hr + r²) = 2hr − h²
∴ Volume v = (1/3)hπ[r² − (h−r)²] = (1/3)πh[2hr − h²] = (1/3)[2h²r − h³]
dv/dh = (1/3)[4hr − 3h²] = 0
⇒ h(4r − 3h) = 0
d²v/dh² = (1/3)[4r − 6h]
At h = 4r/3, d²v/dh² = (1/3)[4r − 8r] < 0
⇒ maximum volume at h = 4r/3
h = 4r/3 = 4
∴ From (1) (h−r)² + b² = r² ⇒ b² = 2hr − h² = 2 × 4r/3 × r − (4r/3)² = 8r²/3 − 16r²/9 = (24 − 16)r²/9 = 8r²/9
⇒ b = 2√2r/3
⇒ 2√2
Curved surface area = πbl = πb√h² + r² = π(2√2)(√16 + 9) = π2√2√25 = π2√2 × 5 = 10√2π
Curved surface area = πb√h² + b² = π(2√2r/3)√(16r²/9) + (8r²/9) = π(2√2r/3)√(24r²/9) = π(2√2r/3)(2√6r/3) = (8√3πr²/9) = 8√3π(9/9) = 8√3π