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JEE Main 2019 · 8 Apr · Shift II

JEE Main 2019 Mathematics question on Application of Derivatives

The height of a right circular cylinder of maximum volume inscribed in a sphere of radius 3 is
  1. A.
    6\sqrt{6}
  2. B.
    232\sqrt{3}
  3. C.
    3\sqrt{3}
  4. D.
    233\dfrac{2}{3}\sqrt{3}
Show answer and solution

Correct option: B

Solution

If the cylinder has height h and base radius r, the sphere constraint gives r² + (h/2)² = 9. Volume V = πr²h = π(9 - h²/4)h. Differentiating and setting dV/dh = 0: 9 - 3h²/4 = 0, so h² = 12, giving h = 2√3.

[JEE Main 2019 · 8 Apr · Shift II]

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