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QJEEM19Ma7d45: JEE Main 2019 Mathematics - Class 12 JEEM19Ma7d45 Solution: The height of a right circular cylinder of maximum volume…

The height of a right circular cylinder of maximum volume inscribed in a sphere of radius 3 is

Multiple Choice Options

  • A 3\sqrt 3
  • B 232\sqrt 3
  • C 6\sqrt 6
  • D 233{2 \over 3} {\sqrt 3}

Step-by-Step Verified Solution

JEE Main 2019 (Online) 8th April Evening Slot Mathematics - Application of Derivatives Question 151 English Explanation
Here r = 3 cos θ\theta

and h2{h \over 2} = 3 sin θ\theta

⇒\Rightarrow h = 6 sin θ\theta

We know, Volume of cylinder

V = π\pi r2 h

⇒\Rightarrow V = π\pi (9cos2 θ\theta )(6 sin θ\theta )

⇒\Rightarrow V = 54 π\pi (sin θ\theta - sin3 θ\theta )

For maxima/minima of volume,

dVdh=0{{dV} \over {dh}} = 0

⇒\Rightarrow cos θ\theta - 3sin2 θ\theta cos θ\theta = 0

⇒\Rightarrow cos θ\theta (1 - 3sin2 θ\theta ) = 0

⇒\Rightarrow cos θ\theta = 0

⇒\Rightarrow θ\theta = π2{\pi \over 2} (not possible)

or 1 - 3sin2 θ\theta = 0

⇒\Rightarrow sin θ\theta = 13{1 \over {\sqrt 3 }}

∴\therefore h = 6 sin θ\theta = 63{6 \over {\sqrt 3 }} = 232\sqrt 3

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