Class 12th   Class 12th Maths

Question 15.

Show that height of the cylinder of greatest volume which can be inscribed in a right circular cone of height $h$ and semi-vertical angle $\alpha$ is one-third that of the cone, and the greatest volume is $\frac{4}{27}\pi h^3{\tan }^2\alpha$.

Solution

Step 1: Geometry of the cone and the inscribed cylinder

Let a cylinder of

• height = $x$

• radius = $r$

be inscribed in a right circular cone of

• height = $h$

• semi-vertical angle = $\alpha$

From the diagram:

The radius at height $x$ from the vertex is proportional to height:

$$\frac{r}{h-x}=\tan \alpha$$

Thus,

$$r=(h-x)\tan \alpha (1)$$

Step 2: Volume of the cylinder

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

Substitute (1):

$$V=\pi {[(h-x)\tan \alpha ]}^{2}x$$
$$V=\pi x(h-x{)}^2{\tan }^2\alpha$$

Let:

$$k=\pi {\tan }^2\alpha$$

Then,

$$V=kx(h-x{)}^2$$

Step 3: Differentiate to find maximum

$$V=k(x{h}^2-2h{x}^2+x^3)$$
$$\frac{dV}{dx}=k(h^2-4hx+3x^2)$$

Set $\frac{dV}{dx}=0$:

$$h^2-4hx+3x^2=0$$

Solve quadratic:

$$3x^2-4hx+h^2=0$$
$$x=\frac{4h\pm \sqrt{16h^2-12h^2}}{6}$$
$$x=\frac{4h\pm 2h}{6}$$

Possible values:

$$x=h\text{or}x=\frac{h}{3}$$

The cylinder cannot have height = $h$ (radius would be 0).

Hence the valid maximum is:

$$\boxed{{\displaystyle x=\frac{h}{3}}}$$

Thus the height of the cylinder of greatest volume is one-third that of the cone.

Step 4: Maximum radius

Use (1):

$$r=(h-x)\tan \alpha =(h-\frac{h}{3})\tan \alpha$$
$$r=\frac{2h}{3}\tan \alpha$$

Step 5: Maximum Volume

$$V_{\max }=\pi r^{2}x$$
$$V_{\max }=\pi {(\frac{2h}{3}\tan \alpha )}^2\left(\frac{h}{3}\right)$$
$$V_{\max }=\pi \left(\frac{4h^2}{9}{\tan }^2\alpha \right)\left(\frac{h}{3}\right)$$
$$V_{\max }=\frac{4}{27}\pi h^3{\tan }^2\alpha$$
$$\boxed{{\displaystyle V_{\max }=\frac{4}{27}\pi h^3{\tan }^2\alpha }}$$

Answers:-

Height of cylinder of greatest volume:

$$\boxed{{\displaystyle \frac{h}{3}}}$$

Greatest possible volume:

$$\boxed{{\displaystyle \frac{4}{27}\pi h^3{\tan }^2\alpha }}$$