Applications of Derivatives
Maxima and Minima
Grade 12
Question:
<p>The maximum volume (in cu.m) of the right circular cone having slant height 3 m is:</p>
<p>\(6\pi\)</p>
<p>\(3\sqrt{3}\,\pi\)</p>
<p>\(\dfrac{4}{3}\,\pi\)</p>
<p>\(2\sqrt{3}\,\pi\)</p>
Step-by-Step Solution
Key Concept: For a cone with fixed slant height, volume V = (1/3)πr²h depends on radius r and height h where r² + h² = 9. Express V as a function of one variable and find its maximum using calculus.
<p><strong>Step 1:</strong> Set up the constraint. For a right circular cone with slant height l = 3 m, radius r, and height h: r² + h² = 9, so h = √(9 - r²)</p><p><strong>Step 2:</strong> Express volume as a function of r: V(r) = (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 · r[2(9 - r²) - r²]/√(9 - r²) = (1/3)π · r(18 - 3r²)/√(9 - r²)</p><p><strong>Step 5:</strong> Set dV/dr = 0: r(18 - 3r²) = 0. Since r > 0, we get 18 - 3r² = 0, so r² = 6, thus r = √6</p><p><strong>Step 6:</strong> Find h: h = √(9 - 6) = √3</p><p><strong>Step 7:</strong> Calculate maximum volume: V = (1/3)π(6)(√3) = 2π√3 cu.m</p><p>∴ Answer: D (2π√3 cu.m)</p>
Correct Answer: D