Class 12th   Class 12th Maths

Question 14.

Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius $R$ is $\frac{2R}{\sqrt{3}}$. Also find the maximum volume.

Solution

Step 1: Understand the geometry

A cylinder is inscribed in a sphere of radius $R$.
Let:

• $h$ = height of the cylinder

• $r$ = radius of the base of the cylinder

From the cross-section, the half-height $h\mathrm{/}2$ and cylinder radius $r$ form a right triangle inside the sphere:

$$r^2+{\left(\frac{h}{2}\right)}^2=R^2(1)$$

Step 2: Write the volume of the cylinder

$$V=\pi r^{2}h$$

Using (1):

$$r^2=R^2-\frac{h^2}{4}$$

So,

$$V(h)=\pi (R^2-\frac{h^2}{4})h$$
$$V(h)=\pi (R^{2}h-\frac{h^3}{4})$$

Step 3: Differentiate to find maximum

$$\frac{dV}{dh}=\pi (R^2-\frac{3h^2}{4})$$

For maximum volume:

$$\frac{dV}{dh}=0$$
$$R^2-\frac{3h^2}{4}=0$$
$$\frac{3h^2}{4}=R^2$$
$$h^2=\frac{4R^2}{3}$$
$$\boxed{{\displaystyle h=\frac{2R}{\sqrt{3}}}}$$

This is the required height.

Step 4: Find the corresponding radius

Using equation (1):

$$r^2=R^2-\frac{h^2}{4}$$

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

$$r^2=R^2-\frac{1}{4}\left(\frac{4R^2}{3}\right)$$
$$r^2=R^2-\frac{R^2}{3}$$
$$r^2=\frac{2R^2}{3}$$

Step 5: Maximum Volume

$$V_{\max }=\pi r^{2}h$$
$$V_{\max }=\pi \left(\frac{2R^2}{3}\right)\left(\frac{2R}{\sqrt{3}}\right)$$
$$V_{\max }=\frac{4\pi R^3}{3\sqrt{3}}$$
$$\boxed{{\displaystyle V_{\max }=\frac{4\pi R^3}{3\sqrt{3}}}}$$

Answers

Height of cylinder of maximum volume:

$$\boxed{{\displaystyle \frac{2R}{\sqrt{3}}}}$$

Maximum volume:

$$\boxed{{\displaystyle V_{\max }=\frac{4\pi R^3}{3\sqrt{3}}}}$$

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