Application Of Derivatives Ques 2
- The height of a right circular cylinder of maximum volume inscribed in a sphere of radius $3$ is
(2019 Main, 8 April II)
(a) $\sqrt{6}$
(b) $2 \sqrt{3}$
(c) $\sqrt{3}$
(d) $\frac{2}{3} \sqrt{3}$
Show Answer
Answer:
Correct Answer: 2.(b)
Solution: Key Idea
(i) Use formula of volume of cylinder, $V=\pi r^2 h$
where, $r=$ radius and $h=$ height
(ii) For maximum or minimum, put first derivative of $V$ equal to zero
Let a sphere of radius $3$ , which inscribed a right circular cylinder having radius $r$ and height is $h$, so
From the figure, $\frac{h}{2}=3 \cos \theta$
$\Rightarrow h=6 \cos \theta $
and $ r=3 \sin \theta $ $\quad$ ……..(i)
$\because$ Volume of cylinder $V=\pi r^2 h$
$ =\pi(3 \sin \theta)^2(6 \cos \theta)=54 \pi \sin ^2 \theta \cos \theta \text {. } $
For maxima or minima, $\frac{d V}{d \theta}=0$
$\Rightarrow 54 \pi\left[2 \sin \theta \cos ^2 \theta-\sin ^3 \theta\right]=0 $
$\Rightarrow \sin \theta\left[2 \cos ^2 \theta-\sin ^2 \theta\right]=0 $
$\Rightarrow \tan ^2 \theta=2 $ $\quad\left[\because \theta \in\left(0, \frac{\pi}{2}\right)\right]$
$\Rightarrow \tan \theta=\sqrt{2} $
$\Rightarrow \sin \theta=\sqrt{\frac{2}{3}} \text { and } \cos \theta=\frac{1}{\sqrt{3}} $ $\quad$ ……..(ii)
From Eqs. (i) and (ii), we get
$ h=6 \frac{1}{\sqrt{3}}=2 \sqrt{3} $
BETA
