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JEE Main 2019 · 9 Jan · Shift I

JEE Main 2019 Mathematics question on Application of Derivatives

The maximum volume (in cu.m) of the right circular cone having slant height 3m is
  1. A.
    43π\frac{4}{3}\pi
  2. B.
    23π2\sqrt{3}\pi
  3. C.
    33π3\sqrt{3}\pi
  4. D.
    6π6\pi
Show answer and solution

Correct option: B

Solution

Let slant height l = 3, radius r, height h = √(9 - r²).

Volume V = (1/3)πr²h = (π/3)r²√(9 - r²).

Maximize V² = (π²/9)r⁴(9 - r²). Differentiate with respect to r²: let u = r², then f(u) = u²(9 - u). f'(u) = 18u - 3u² = 3u(6 - u) = 0, giving r² = 6, so h = √3.

V = (π/3)(6)(√3) = 2√3π.

[JEE Main 2019 · 9 Jan · Shift I]

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