Question

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 greatest volume of cylinder is
$4 \div 27\pi {h}^{3} \ { \tan( \alpha ) }^{2}$
​

Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that

↝ Radius of cylinder = x

and

↝ Height of cylinder = y

and

↝ Volume of cylinder = V

Given that,

↝ Height of cone = h and semi - vertical angle is alpha.

So, from figure, we get

↝ Height of cylinder, y = h - a

$\bf \implies\:y = h - x \: cot \alpha - - - (1)$

Now, we know,

$\rm :\longmapsto\:\boxed{ \tt{ \: Volume_{cylinder} = \pi \: {r}^{2} \: h \: }}$

where,

r is radius of cylinder

h is height of cylinder

So, on substituting the values, we get

$\rm :\longmapsto\:V = \pi \: {x}^{2} (h - xcot \alpha )$

can be further rewritten as

$\rm :\longmapsto\:V = \pi \: (h{x}^{2} - {x}^{3} cot \alpha )$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}V =\dfrac{d}{dx} \pi \: (h{x}^{2} - {x}^{3} cot \alpha )$

$\rm :\longmapsto\:\dfrac{dV}{dx} = \pi \: (2hx - 3 {x}^{2}cot \alpha )$

For maxima or minima,

$\rm :\longmapsto\:\dfrac{dV}{dx} = 0$

$\rm :\longmapsto\:\pi \: (2xh - 3 {x}^{2} cot \alpha ) = 0$

$\rm :\longmapsto\:2hx = {3x}^{2}cot \alpha$

$\rm :\longmapsto\:2h= {3x}^{}cot \alpha$

$\bf\implies \:\boxed{ \tt{ \: x = \frac{2h}{3cot \alpha } \: }}$

Now, again,

$\rm :\longmapsto\:\dfrac{dV}{dx} = \pi \: (2hx - 3 {x}^{2}cot \alpha )$

So, on differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{ {d}^{2} V}{ {dx}^{2} } = \pi(2h - 6xcot \alpha )$

On substituting the value of x, we get

$\rm :\longmapsto\:\dfrac{ {d}^{2} V}{ {dx}^{2} } = \pi\bigg(2h - 6cot \alpha \times \dfrac{2h}{3cot \alpha }\bigg )$

$\rm :\longmapsto\:\dfrac{ {d}^{2} V}{ {dx}^{2} } = \pi\bigg(2h - 4h\bigg )$

$\rm :\longmapsto\:\dfrac{ {d}^{2} V}{ {dx}^{2} } = \pi\bigg(- 2h\bigg )$

$\rm :\longmapsto\:\dfrac{ {d}^{2} V}{ {dx}^{2} } < 0$

$\bf\implies \:V \: is \: maximum \: at \: x = \dfrac{2h}{3cot \alpha }$

Part - 1

Height of Cylinder

As from above, Height of cylinder, y

$\rm :\longmapsto\:\:y = h - x \: cot \alpha$

$\rm :\longmapsto\:\:y = h - \: cot \alpha \times \dfrac{2h}{3cot \alpha }$

$\rm :\longmapsto\:\:y = h - \: \dfrac{2h}{3 }$

$\rm :\longmapsto\:\:y = \: \dfrac{3h - 2h}{3 }$

$\rm :\longmapsto\:\:y = \: \dfrac{h}{3}$

Hence,

• Height of cylinder is one third of height of cone.

Part - 2

Volume of cylinder

From above,

Volume of cylinder V is given by

$\rm :\longmapsto\:V = \pi \: {x}^{2} (h - xcot \alpha )$

On substituting the value of x, we get

$\rm :\longmapsto\:V = \pi \: {\bigg[\dfrac{2h}{3cot \alpha } \bigg]}^{2} \bigg(h - cot \alpha \bigg[\dfrac{2h}{3cot \alpha } \bigg]\bigg)$

$\rm :\longmapsto\:V = \pi \: {\bigg[\dfrac{4 {h}^{2} }{9cot^{2} \alpha } \bigg]}^{} \bigg(h - \dfrac{2h}{3} \bigg)$

$\rm :\longmapsto\:V = \pi \: {\bigg[\dfrac{4 {h}^{2} }{9cot^{2} \alpha } \bigg]}^{} \bigg( \dfrac{3h - 2h}{3}\bigg)$

$\rm :\longmapsto\:V = \pi \: {\bigg[\dfrac{4 {h}^{2} }{9cot^{2} \alpha } \bigg]}^{} \bigg( \dfrac{h}{3}\bigg)$

$\bf\implies \:V = \dfrac{4}{27}\pi \: {h}^{3} \: {tan}^{2} \alpha$

Additional Information

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$