Name: Increasing Functions
Uploaded: 2019-07-19
Duration: 3 min 59 s
Description: The GDP of India is either increasing or decreasing, depending on a variety of factors. Let’s understand what an increasing function is, by using derivatives...

Question

A cylinder is inscribed in a sphere of radius

R

. Prove that its volume is maximum if its height is

\frac{2 R}{\sqrt{3}}

.

Accepted Answer

We were given that the radius of the sphere is R.
Let the radius of the cylinder be r and the height of the cylinder be h.
Let us consider the formula for the volume of cylinder,
Volume of cylinder $= π \times {(r a d i u s)}^{2} \times (h e i g h t)$
Using the above formula, we can write the volume of the cylinder as,
$V = π r^{2} h$
Now, as the cylinder is inscribed inside a sphere, the centre of the sphere is the same as the center of the cylinder.

So, now let us take a look at the figure below.

As we see above triangle is right-angled. Hence, applying the Pythagoras theorem, we get
$\Rightarrow r^{2} + (\frac{h^{2}}{2}) = R^{2}$ $\Rightarrow r^{2} + \frac{h^{2}}{4} = R^{2}$ $\Rightarrow r^{2} = R^{2} - \frac{h^{2}}{4}$
Now let us substitute the value of $r$ in the volume. Then, we get
$\Rightarrow V = π r^{2} h$

$\Rightarrow V = π (R^{2} - \frac{h^{2}}{4}) h$ $\Rightarrow V = π R^{2} h - π \frac{h^{3}}{4}$
Now let us discuss the concept of optimization.

A function $f (x)$ is said to achieve extremum that is maximum or minimum if it satisfies the condition,
$f' (x) = 0$
As we need to find the maximum value of volume, let us differentiate the volume with respective to $h$ .
$\Rightarrow \frac{d V}{d h} = \frac{d}{d h} (π R^{2} h - π \frac{h^{3}}{4})$ $\Rightarrow \frac{d V}{d h} = \frac{d}{d h} (π R^{2} h) - \frac{d}{d h} (π \frac{h^{3}}{4})$ $\Rightarrow \frac{d V}{d h} = π R^{2} - π \frac{3 h^{2}}{4}$
When the volume is maximum, $\frac{d V}{d h} = 0$ .

So, equating the value of $\frac{d V}{d h}$ to zero.
$\Rightarrow \frac{d V}{d h} = π R^{2} - π \frac{3 h^{2}}{4} = 0$

$\Rightarrow π R^{2} = π \frac{3 h^{2}}{4}$

$\Rightarrow R^{2} = \frac{3 h^{2}}{4}$

$\Rightarrow h^{2} = \frac{4 R^{2}}{3}$

$\Rightarrow h = \frac{2 R}{\sqrt{3}}$
So, we get the height of cylinder as $h = \frac{2 R}{\sqrt{3}}$ .

A cylinder is inscribed in a sphere of radius $R$ . Prove that its volume is maximum if its height is $\frac{2 R}{\sqrt{3}}$ .

Important Questions on Application of Derivatives

Learn and Prepare for any exam you want !

A cylinder is inscribed in a sphere of radius R. Prove that its volume is maximum if its height is 2R3.

Important Questions on Application of Derivatives

Learn and Prepare for any exam you want !

A cylinder is inscribed in a sphere of radius $R$ . Prove that its volume is maximum if its height is $\frac{2 R}{\sqrt{3}}$ .