MHT CET · Maths · Application of Derivatives
The height of right circular cylinder of maximum volume inscribed in a sphere of diameter \(2 a\) is
- A \(2 \sqrt{3} a\)
- B \(\sqrt{3} a\)
- C \(\frac{2 a}{\sqrt{3}}\)
- D \(\frac{a}{\sqrt{3}}\)
Answer & Solution
Correct Answer
(C) \(\frac{2 a}{\sqrt{3}}\)
Step-by-step Solution
Detailed explanation
Let the radius and height of the cylinder are and \(h\), respectively.

In \(\triangle A O M\)
\(
\begin{array}{l} \therefore r^{2}+\left(\frac{h^{2}}{4}\right)=a^{2} \\
h^{2}=4\left(a^{2}-r^{2}\right)
\end{array}
\)
Now, \(V=\pi r^{2} h=\pi\left(a^{2} h-\frac{1}{4} h^{3}\right)\)
For max or min,
\(\frac{d V}{d h} =\pi\left(a^{2}-\frac{3}{4} h^{2}\right)=0 \)
\( \Rightarrow h =\left(\frac{2}{\sqrt{3}}\right) a\)
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 \(\triangle A O M\)
\(
\begin{array}{l} \therefore r^{2}+\left(\frac{h^{2}}{4}\right)=a^{2} \\
h^{2}=4\left(a^{2}-r^{2}\right)
\end{array}
\)
Now, \(V=\pi r^{2} h=\pi\left(a^{2} h-\frac{1}{4} h^{3}\right)\)
For max or min,
\(\frac{d V}{d h} =\pi\left(a^{2}-\frac{3}{4} h^{2}\right)=0 \)
\( \Rightarrow h =\left(\frac{2}{\sqrt{3}}\right) a\)
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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