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The height of right circular cylinder of maximum volume inscribed in a sphere of diameter $2 a$ is
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2568 Upvotes
Verified Answer
The correct answer is:
$\frac{2 a}{\sqrt{3}}$
Let the radius and height of the cylinder are and $h$, respectively.
In $\triangle A O M$
$\therefore$
$$
\begin{array}{l}
r^{2}+\left(\frac{h^{2}}{4}\right)=a^{2} \\
h^{2}=4\left(a^{2}-r^{2}\right)
\end{array}
$$
Now, $V=\pi r^{2} h=\pi\left(a^{2} h-\frac{1}{4} h^{3}\right)$
For max or min,
$$
\begin{aligned}
& \frac{d V}{d h} &=\pi\left(a^{2}-\frac{3}{4} h^{2}\right)=0 \\
\Rightarrow & h &=\left(\frac{2}{\sqrt{3}}\right) a
\end{aligned}
$$
Now, $\quad \frac{d^{2} V}{d h^{2}}=-\frac{6 h}{4} < 0$
So, $V$ is maximum at $h=\frac{2 a}{\sqrt{3}}$.
In $\triangle A O M$
$\therefore$
$$
\begin{array}{l}
r^{2}+\left(\frac{h^{2}}{4}\right)=a^{2} \\
h^{2}=4\left(a^{2}-r^{2}\right)
\end{array}
$$
Now, $V=\pi r^{2} h=\pi\left(a^{2} h-\frac{1}{4} h^{3}\right)$
For max or min,
$$
\begin{aligned}
& \frac{d V}{d h} &=\pi\left(a^{2}-\frac{3}{4} h^{2}\right)=0 \\
\Rightarrow & h &=\left(\frac{2}{\sqrt{3}}\right) a
\end{aligned}
$$
Now, $\quad \frac{d^{2} V}{d h^{2}}=-\frac{6 h}{4} < 0$
So, $V$ is maximum at $h=\frac{2 a}{\sqrt{3}}$.
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