Class 12 Applications Of Derivatives · Very Long Answer5 Marksgeometric_optimization

Class 12 Maths · Applications Of Derivatives · Very Long Answer #72

Prove that a right circular cone with fixed slant height l>0 has greatest volume when its semi-vertical angle is arccos(1/√3).

✅ Step-by-Step Solution

Final Answer: θ=arccos(1/√3).

For 0≤θ≤π/2 the radius is l sinθ and height l cosθ.
V(θ)=πl³sin²θ cosθ/3.
V′=(πl³/3)sinθ(3cos²θ−1); the interior zero satisfies cosθ=1/√3.
The derivative is positive before this angle and negative after it; both degenerate endpoint volumes are zero, proving the global maximum.
Thus θ=arccos(1/√3), with height l/√3 and radius l√(2/3).

[1 mark] For 0≤θ≤π/2 the radius is l sinθ and height l cosθ.
[1 mark] V(θ)=πl³sin²θ cosθ/3.
[1 mark] V′=(πl³/3)sinθ(3cos²θ−1); the interior zero satisfies cosθ=1/√3.
[1 mark] The derivative is positive before this angle and negative after it; both degenerate endpoint volumes are zero, proving the global maximum.
[1 mark] Thus θ=arccos(1/√3), with height l/√3 and radius l√(2/3).

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