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JEE Mains Previous Paper 1 (Held On: 08 Apr 2019 Shift 2)

Option 3 : \(2\sqrt 3\)

JEE Mains Previous Paper 1 (Held On: 12 Apr 2019 Shift 2)

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Let radius of base and height of cylinder be r and h respectively.

\(\therefore {r^2} + \frac{{{h^2}}}{4} = 9\) ----(1)

Now, volume of cylinder, V = πr^{2}h

Substitute the value of r^{2} from equation (1)

\(V = \pi h\left( {9 - \frac{{{h^2}}}{4}} \right)\)

\(⇒ V = 9\pi h - \frac{\pi }{4}{h^3}\)

Differentiating with respect to h

\(\frac{{dV}}{{dh}} = 9\pi - \frac{3}{4}\pi {h^2}\) ----(2)

For maxima/minima,

\(\frac{{dV}}{{dh}} = 0 ⇒ 9\pi - \frac{3}{4}\pi {h^2} = 0\)

\(⇒ 9 = \frac{3}{4}{h^2} ⇒ {h^2} = \frac{{9 \times 4}}{3} = 12\)

\(h = \sqrt {12}\)

Differentiating with respect to h equation (2)

and \(\frac{{{d^2}V}}{{d{h^2}}} = - \frac{3}{2}\pi h\)

\(\therefore {\left( {\frac{{{d^2}V}}{{d{h^2}}}} \right)_{h = \sqrt {12} }} < 0\)

⇒ Volume is maximum when \(h = \sqrt {12} = 2\sqrt 3\).