# Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is${\tan ^{ - 1}}\sqrt 2 $.

Answer

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Hint:Express volume in terms of slant height and height , then differentiate that expression wrt height because slant height here is given constant. After getting a certain relation use basic trigonometry.

Let, height of cone$ = h$

Radius of cone$ = r$

Slant height$ = l$

Also the slant height is given by ${l^2} = {r^2} + {h^2}$

As we know volume$\left( V \right)$ of the cone$ = \dfrac{\pi }{3}{r^2}h$

$

= \dfrac{\pi }{3}\left( {{l^2} - {h^2}} \right)h \\

= \dfrac{\pi }{3}\left( {{l^2}h - {h^3}} \right) \\

$

Now we have to maximize the volume and we know that volume is maximum if height is maximum so, differentiate the volume w.r.t .$h$ and equate to zero

$

V = \dfrac{\pi }{3}\left( {{l^2}h - {h^3}} \right) \\

\dfrac{{dV}}{{dh}} = \dfrac{\pi }{3}\left( {{l^2} - 3{h^2}} \right) = 0 \\

\Rightarrow {l^2} = 3{h^2} \\

\Rightarrow l = \sqrt 3 h \\

$

Let, $\theta $be the semi vertical angle

$ \Rightarrow \cos \theta = \dfrac{{{\text{height}}}}{{{\text{slant height}}}} = \dfrac{h}{l} = \dfrac{h}{{h\sqrt 3 }} = \dfrac{1}{{\sqrt 3 }}$

Now we know that ${\tan ^2}\theta = {\sec ^2}\theta - 1,{\text{ }}\sec \theta = \dfrac{1}{{\cos \theta }}$

\[

{\tan ^2}\theta = \dfrac{1}{{{{\cos }^2}\theta }} - 1 = \dfrac{{1 - {{\cos }^2}\theta }}{{{{\cos }^2}\theta }} \\

\Rightarrow \tan \theta = \dfrac{{\sqrt {1 - {{\cos }^2}\theta } }}{{\cos \theta }} = \dfrac{{\sqrt {1 - \dfrac{1}{3}} }}{{\dfrac{1}{{\sqrt 3 }}}} = \dfrac{{\dfrac{{\sqrt 2 }}{{\sqrt 3 }}}}{{\dfrac{1}{{\sqrt 3 }}}} = \sqrt 2 \\

\Rightarrow \theta = {\tan ^{ - 1}}\sqrt 2 \\

\]

Hence Proved

Note: - In such types of problems the key concept we have to remember is that for maximize the volume we have to maximize the height in terms of slant height because slant height is given in this problem, if radius is given than we have to maximize the height in terms of radius by applying maxima and minima concept which is stated above, then by using some basic trigonometric properties which is stated above we will get the required answer.

Let, height of cone$ = h$

Radius of cone$ = r$

Slant height$ = l$

Also the slant height is given by ${l^2} = {r^2} + {h^2}$

As we know volume$\left( V \right)$ of the cone$ = \dfrac{\pi }{3}{r^2}h$

$

= \dfrac{\pi }{3}\left( {{l^2} - {h^2}} \right)h \\

= \dfrac{\pi }{3}\left( {{l^2}h - {h^3}} \right) \\

$

Now we have to maximize the volume and we know that volume is maximum if height is maximum so, differentiate the volume w.r.t .$h$ and equate to zero

$

V = \dfrac{\pi }{3}\left( {{l^2}h - {h^3}} \right) \\

\dfrac{{dV}}{{dh}} = \dfrac{\pi }{3}\left( {{l^2} - 3{h^2}} \right) = 0 \\

\Rightarrow {l^2} = 3{h^2} \\

\Rightarrow l = \sqrt 3 h \\

$

Let, $\theta $be the semi vertical angle

$ \Rightarrow \cos \theta = \dfrac{{{\text{height}}}}{{{\text{slant height}}}} = \dfrac{h}{l} = \dfrac{h}{{h\sqrt 3 }} = \dfrac{1}{{\sqrt 3 }}$

Now we know that ${\tan ^2}\theta = {\sec ^2}\theta - 1,{\text{ }}\sec \theta = \dfrac{1}{{\cos \theta }}$

\[

{\tan ^2}\theta = \dfrac{1}{{{{\cos }^2}\theta }} - 1 = \dfrac{{1 - {{\cos }^2}\theta }}{{{{\cos }^2}\theta }} \\

\Rightarrow \tan \theta = \dfrac{{\sqrt {1 - {{\cos }^2}\theta } }}{{\cos \theta }} = \dfrac{{\sqrt {1 - \dfrac{1}{3}} }}{{\dfrac{1}{{\sqrt 3 }}}} = \dfrac{{\dfrac{{\sqrt 2 }}{{\sqrt 3 }}}}{{\dfrac{1}{{\sqrt 3 }}}} = \sqrt 2 \\

\Rightarrow \theta = {\tan ^{ - 1}}\sqrt 2 \\

\]

Hence Proved

Note: - In such types of problems the key concept we have to remember is that for maximize the volume we have to maximize the height in terms of slant height because slant height is given in this problem, if radius is given than we have to maximize the height in terms of radius by applying maxima and minima concept which is stated above, then by using some basic trigonometric properties which is stated above we will get the required answer.

Last updated date: 28th Sep 2023

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