Applications of Derivatives
Maxima and Minima
Grade 12
Question:
<p>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</p>
<p>\(6\sqrt{2}\,\pi\)</p>
<p>\(6\sqrt{3}\,\pi\)</p>
<p>\(8\sqrt{2}\,\pi\)</p>
<p>\(8\sqrt{3}\,\pi\)</p>
Step-by-Step Solution
Key Concept: For a cone inscribed in a sphere, express the cone's volume in terms of its height using the sphere's constraint, then optimize by differentiating with respect to height to find maximum volume.
<p><strong>Step 1:</strong> Set up the constraint. For a cone inscribed in a sphere of radius R = 3, let h be the height and r be the base radius. The sphere's center lies on the cone's axis. If the apex is at distance (R-h') from center and base at distance d from center, then r² + d² = R². For a cone with apex on sphere: h = R + d, and r² = R² - d².</p><p><strong>Step 2:</strong> Express volume as V = (1/3)πr²h = (1/3)π(R² - d²)(R + d). Let d vary from -R to R. Substituting R = 3: V = (1/3)π(9 - d²)(3 + d).</p><p><strong>Step 3:</strong> Expand: V = (1/3)π(27 + 9d - 3d² - d³). Differentiate: dV/dd = (1/3)π(9 - 6d - 3d²). Setting dV/dd = 0: 3d² + 6d - 9 = 0, so d² + 2d - 3 = 0, giving (d+3)(d-1) = 0. Thus d = 1 (taking valid solution).</p><p><strong>Step 4:</strong> When d = 1: r² = 9 - 1 = 8, so r = 2√2. Height h = 3 + 1 = 4. Slant height l = √(h² + r²) = √(16 + 8) = √24 = 2√6.</p><p><strong>Step 5:</strong> Curved surface area = πrl = π(2√2)(2√6) = 4π√12 = 4π(2√3) = 8π√3 cm².</p><p>∴ Answer: D</p>
Correct Answer: D