Question:

Express $S^2 = \pi^2 r^4 + \dfrac{9V^2}{r^2}$. [1.5 Marks]
$f'(r) = 4\pi^2 r^3 - \dfrac{18V^2}{r^3} = 0 \Rightarrow 4\pi^2 r^6 = 18V^2$. [1.5 Marks]
Substitute $V = \frac{1}{3}\pi r^2 h \Rightarrow 4\pi^2 r^6 = 2\pi^2 r^4 h^2 \Rightarrow h = \sqrt{2} r$. Proved! [2.0 Marks]

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🎯 Official CBSE Marking Scheme:
Setting up curved surface area squared function: 1.5 Marks
Differentiating and solving critical relation: 1.5 Marks
Proving $h = \sqrt{2} r$: 2.0 Marks