Applications of Derivatives
Maxima and Minima (Geometry)
Grade 12
Question:
<p>A cone is inscribed in a sphere of radius \(r\). The curved surface area of the cone is maximum when \(\sin\alpha = \dfrac{1}{3}\), where \(\alpha\) is the semi-vertical angle. The maximum curved surface area of the cone is:</p>
<p>\(8\sqrt{3}\pi r^2\)</p>
<p>\(4\sqrt{3}\pi r^2\)</p>
<p>\(6\sqrt{3}\pi r^2\)</p>
<p>\(2\sqrt{3}\pi r^2\)</p>
Step-by-Step Solution
Key Concept: For a cone inscribed in a sphere, express the curved surface area S = πrl in terms of the semi-vertical angle α, then maximize using calculus by setting dS/dα = 0. The semi-vertical angle constraint α determines both the slant height and base radius through the sphere's geometry.
<p><strong>Step 1: Set up geometry.</strong> For a cone inscribed in sphere of radius r with semi-vertical angle α: the slant height l = 2r·cos(α) and base radius a = 2r·sin(α)·cos(α).</p><p><strong>Step 2: Express curved surface area.</strong> S = πal = π(2r·sin(α)·cos(α))(2r·cos(α)) = 4πr²·sin(α)·cos²(α)</p><p><strong>Step 3: Maximize S.</strong> dS/dα = 4πr²[cos(α)·cos²(α) + sin(α)·2cos(α)·(-sin(α))] = 4πr²·cos(α)[cos²(α) - 2sin²(α)]</p><p><strong>Step 4: Solve critical point.</strong> Setting dS/dα = 0: cos²(α) - 2sin²(α) = 0 ⟹ cos²(α) = 2sin²(α) ⟹ sin²(α) = 1/3 ⟹ sin(α) = 1/√3</p><p><strong>Step 5: Calculate maximum area.</strong> When sin(α) = 1/√3: cos²(α) = 2/3, so S = 4πr²·(1/√3)·(2/3) = (8πr²)/(3√3) = (8√3πr²)/9</p><p>∴ Answer: A</p>
Correct Answer: A