Question

Find the max.value of the total surface of a right circular cylinder which can be inscribed in a sphere of radius a.

• A. $\pi a^2(1+1/\sqrt{5})$
• B. $\pi /2a^2(1+\sqrt{5})$
• C. $\pi a^2(1+\sqrt{5})$
• D. $\pi a^2(2+\sqrt{5})$

Answer 1

The correct option is C $\pi a^2(1+\sqrt{5})$

$S=2\pi r^2+2\pi rh.$ $r=a\cos \theta andh/2=a\sin theta$

$\therefore h=2a\sin \theta$ $\therefore s=2\pi .(a^2{\cos }^2\theta +a\cos \theta .2a\sin \theta )$

$S=2\pi a^2({\cos }^2\theta +\sin 2\theta ).$ $ds/d\theta =2^2[-2\cos \theta \sin \theta +2\cos 2\theta ]$ $=2\pi a^2[2\cos 2\theta -\sin 2\theta ]=0$

$\therefore \tan 2\theta =2or\cos 2\theta =\frac{1}{\sqrt{\left(5\right)}}$ $\sin 2\theta =\frac{2}{\sqrt{\left(5\right)}}$ ${\frac{d^{2}S}{d\theta }}^2=2\pi [-4\sin 2\theta -2\cos 2\theta ]=$-ive and hence max. when $\tan 2\theta =2$

$\therefore S=2\pi a^2({\cos }^2\theta +\sin 2\theta )$ $=2\pi a^2[\frac{1+\cos 2\theta }{2}+\sin 2\theta ]$ $=2\pi a^2[\frac{1}{2}+\frac{1}{2\sqrt{\left(5\right)}}+\frac{2}{\left(5\right)}]$ $=2\pi a^2[\frac{1}{2}+\frac{5}{2\sqrt{\left(5\right)}}]=\pi a^2(1+\sqrt{5})$