UPSC 2014 Maths Optional Paper 1 Q3a — Step-by-Step Solution

15 marks · Section A

Maxima and minima of single-variable functions · Calculus · asked 7× in 13 yrs · Read the full method →

Question

Find the height of the cylinder of maximum volume that can be inscribed in a sphere of radius $a$ .

Single-variable optimisation; standard inscribed-cylinder calculus problem.

Strategy. By symmetry, place the cylinder axis along the sphere’s diameter. Express volume in terms of one parameter (height); maximise.

Let the cylinder have full height $2h$ (so half-height $h$ ) and base radius $r$ . By symmetry the axis passes through the sphere’s centre.

The top edge of the cylinder is on the sphere: $r^{2}+h^{2}=a^{2}$ , hence $r^{2}=a^{2}-h^{2}$ .

Volume:

V(h)=\pi r^{2}\cdot 2h=2\pi h(a^{2}-h^{2}).

\frac{dV}{dh}=2\pi(a^{2}-3h^{2}).

Set $dV/dh=0$ : $h^{2}=a^{2}/3$ , $h=a/\sqrt 3$ .

Second-derivative test:

\frac{d^{2}V}{dh^{2}}=-12\pi h<0\quad\text{at }h=a/\sqrt 3,

so this is a maximum.

Total height = $2h$ :

\boxed{\;\text{Height}=\dfrac{2a}{\sqrt 3}=\dfrac{2a\sqrt 3}{3}.\;}