Question

A conical vessel is to be prepared out of a circular sheet of tin of unit radius. How much sectorial area is to be removed from the sheet so that the vessel has maximum volume?

Accepted Answer

Radius of the sheet $(R)$ $= 1$ unit

After removing the sector, when conical vessel is formed, radius of circle will become slant height $(l)$ of cone.

Therefore, $l = 1$ unit

Circumference of circle after removing sheet $= 2 π R - R θ = 2 π - θ$

Let the radius of the cone be $r$ , then $2 π r = 2 π - θ$

$\Rightarrow r = \frac{2 π - θ}{2 π} \to (1)$ [Here $r$ is radius of a cone]

Height of the cone $h = \sqrt{1^{2} - r^{2}}$

Therefore, Volume of the cone is $V = \frac{1}{3} π (1 - h^{2}) h$

$\Rightarrow V = \frac{1}{3} π (h - h^{3})$

So, $\frac{d V}{d h} = \frac{1}{3} π (1 - 3 h^{2})$

For maxima or minima, $\frac{d V}{d h} = 0$

$\Rightarrow \frac{1}{3} π (1 - 3 h^{2}) = 0$

$\Rightarrow 3 h^{2} - 1 = 0$

$\Rightarrow h = \pm \frac{1}{\sqrt{3}}$

As height $(h)$ cannot be negative, thus the required value of $h$ for maximum volume should be $h = \frac{1}{\sqrt{3}}$ and $r = \frac{\sqrt{2}}{\sqrt{3}}$

Substituting $r = \frac{\sqrt{2}}{\sqrt{3}}$ in $(1)$ , we get: $\frac{2 π - θ}{2 π} = \frac{\sqrt{2}}{\sqrt{3}}$

$\Rightarrow 2 π - θ = \frac{2 \sqrt{2} π}{\sqrt{3}}$

$\Rightarrow θ = 2 π - \frac{2 \sqrt{2} π}{\sqrt{3}}$

$\Rightarrow θ = \frac{2 π}{\sqrt{3}} (\sqrt{3} - \sqrt{2})$

Now, the area of the sector removed $= \frac{1}{2} θ \cdot R^{2}$

$= \frac{1}{2} \cdot \frac{2 π}{\sqrt{3}} (\sqrt{3} - \sqrt{2}) \cdot 1^{2}$

$= \frac{π (\sqrt{3} - \sqrt{2})}{\sqrt{3}}$

Hence, the area of the sector removed is $\frac{π (\sqrt{3} - \sqrt{2})}{\sqrt{3}}$ sq.unit.

A conical vessel is to be prepared out of a circular sheet of tin of unit radius. How much sectorial area is to be removed from the sheet so that the vessel has maximum volume?

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