<p>A cone has slant height \(l = 3\) m. The maximum volume (in cu. m) of the cone is:</p>
Step-by-Step Solution
Key Concept: For a cone with fixed slant height l, express volume V in terms of a single variable (base radius r) using the constraint l² = r² + h², then find maximum using calculus.
<p><strong>Step 1:</strong> Set up the constraint. For a cone with slant height l = 3 and base radius r, height h is given by: l² = r² + h² → 9 = r² + h² → h = √(9 - r²)</p><p><strong>Step 2:</strong> Express volume as a function of r alone: V = (1/3)πr²h = (1/3)πr²√(9 - r²)</p><p><strong>Step 3:</strong> Find dV/dr. Using product rule: dV/dr = (1/3)π[2r√(9 - r²) + r² · (-r/√(9 - r²))] = (1/3)π[2r√(9 - r²) - r³/√(9 - r²)]</p><p><strong>Step 4:</strong> Simplify: dV/dr = (1/3)π · [2r(9 - r²) - r³]/√(9 - r²) = (1/3)π · r[18 - 3r²]/√(9 - r²)</p><p><strong>Step 5:</strong> Set dV/dr = 0: r(18 - 3r²) = 0 → r = 0 or r² = 6 → r = √6 (taking positive value)</p><p><strong>Step 6:</strong> When r = √6, h = √(9 - 6) = √3</p><p><strong>Step 7:</strong> Maximum volume: V = (1/3)π(6)(√3) = 2√3π cu. m</p><p>∴ Answer: A</p>
Correct Answer: A