Answer

Verified

405.9k+ views

Hint: First find the surface area of the cone. After that, take $l=\sqrt{{{h}^{2}}+{{r}^{2}}}$ and compare to get the volume. Then for maximum volume, $\dfrac{dV}{dr}=0$, from this you will get the value of $S$ and substitute in $l$, you will get the value of $l$ in terms of $r$ and also you will get the value of $h$. The take $\tan $ of semi vertical angle and you will get the answer.

Complete step-by-step answer:

Now the surface area of the cone will be $S=\pi {{r}^{2}}+\pi rl$ where r is the radius and l is the slant height of the cone.

Now $l=\dfrac{S-\pi {{r}^{2}}}{\pi r}$……….. (1)

Also from above figure we know that ${{h}^{2}}+{{r}^{2}}={{l}^{2}}$, where $h$ is the vertical height of the cone.

So we get, $l=\sqrt{{{h}^{2}}+{{r}^{2}}}$ ……….(2)

From (1) and (2), we get,

$\sqrt{{{h}^{2}}+{{r}^{2}}}=\dfrac{S-\pi {{r}^{2}}}{\pi r}$

Squaring both sides we get,

\[\begin{align}

& {{h}^{2}}+{{r}^{2}}={{\left( \dfrac{S-\pi {{r}^{2}}}{\pi r} \right)}^{2}} \\

& {{h}^{2}}=\left( \dfrac{{{S}^{2}}+{{\pi }^{2}}{{r}^{4}}-2\pi S{{r}^{2}}}{{{\pi }^{2}}{{r}^{2}}} \right)-{{r}^{2}} \\

& {{h}^{2}}=\left( \dfrac{{{S}^{2}}+{{\pi }^{2}}{{r}^{4}}-2\pi S{{r}^{2}}-{{\pi }^{2}}{{r}^{4}}}{{{\pi }^{2}}{{r}^{2}}} \right) \\

& {{h}^{2}}=\left( \dfrac{{{S}^{2}}-2\pi S{{r}^{2}}}{{{\pi }^{2}}{{r}^{2}}} \right) \\

& h=\dfrac{\sqrt{{{S}^{2}}-2\pi S{{r}^{2}}}}{\pi r} \\

\end{align}\]

Now volume of cone V$=\dfrac{1}{3}\pi {{r}^{2}}h=\dfrac{1}{3}\pi {{r}^{2}}\dfrac{\sqrt{{{S}^{2}}-2\pi S{{r}^{2}}}}{\pi r}$

V$=\dfrac{r}{3}\sqrt{{{S}^{2}}-2\pi S{{r}^{2}}}$

For maximum volume, $\dfrac{dV}{dr}=0$.

$\begin{align}

& \dfrac{1}{3}\sqrt{{{S}^{2}}-2\pi S{{r}^{2}}}+\dfrac{r}{3}\dfrac{-4\pi Sr}{2\sqrt{{{S}^{2}}-2\pi S{{r}^{2}}}}=0 \\

& \dfrac{1}{3}\sqrt{{{S}^{2}}-2\pi S{{r}^{2}}}=\dfrac{r}{3}\dfrac{2\pi Sr}{\sqrt{{{S}^{2}}-2\pi S{{r}^{2}}}} \\

& {{S}^{2}}=4\pi S{{r}^{2}} \\

\end{align}$

we know that, $S\ne 0$.

So $S=4\pi {{r}^{2}}$

Now taking (1) and substituting the $S=4\pi {{r}^{2}}$ we get,

$l=\dfrac{4\pi {{r}^{2}}-\pi {{r}^{2}}}{\pi r}=3r$

So $l=\sqrt{{{h}^{2}}+{{r}^{2}}}$

$\begin{align}

& 9{{r}^{2}}={{h}^{2}}+{{r}^{2}} \\

& h=2\sqrt{2}r \\

\end{align}$

Now, let $\theta $ be the semi-vertical angle of the cone.

$\begin{align}

& \tan \theta =\dfrac{r}{h}=\dfrac{r}{2\sqrt{2}r} \\

& \tan \theta =\dfrac{1}{2\sqrt{2}} \\

\end{align}$

So $\theta ={{\tan }^{-1}}\left( \dfrac{1}{2\sqrt{2}} \right)$

Hence proved.

Note: This question deals with maxima hence differentiation is involved. Few relations like the one between r, l & h are determined based on the geometry of the figure. For finding the condition for maximum volume, volume is differentiated wrt the radius and equated to 0. Avoid mistakes while differentiating.

Complete step-by-step answer:

Now the surface area of the cone will be $S=\pi {{r}^{2}}+\pi rl$ where r is the radius and l is the slant height of the cone.

Now $l=\dfrac{S-\pi {{r}^{2}}}{\pi r}$……….. (1)

Also from above figure we know that ${{h}^{2}}+{{r}^{2}}={{l}^{2}}$, where $h$ is the vertical height of the cone.

So we get, $l=\sqrt{{{h}^{2}}+{{r}^{2}}}$ ……….(2)

From (1) and (2), we get,

$\sqrt{{{h}^{2}}+{{r}^{2}}}=\dfrac{S-\pi {{r}^{2}}}{\pi r}$

Squaring both sides we get,

\[\begin{align}

& {{h}^{2}}+{{r}^{2}}={{\left( \dfrac{S-\pi {{r}^{2}}}{\pi r} \right)}^{2}} \\

& {{h}^{2}}=\left( \dfrac{{{S}^{2}}+{{\pi }^{2}}{{r}^{4}}-2\pi S{{r}^{2}}}{{{\pi }^{2}}{{r}^{2}}} \right)-{{r}^{2}} \\

& {{h}^{2}}=\left( \dfrac{{{S}^{2}}+{{\pi }^{2}}{{r}^{4}}-2\pi S{{r}^{2}}-{{\pi }^{2}}{{r}^{4}}}{{{\pi }^{2}}{{r}^{2}}} \right) \\

& {{h}^{2}}=\left( \dfrac{{{S}^{2}}-2\pi S{{r}^{2}}}{{{\pi }^{2}}{{r}^{2}}} \right) \\

& h=\dfrac{\sqrt{{{S}^{2}}-2\pi S{{r}^{2}}}}{\pi r} \\

\end{align}\]

Now volume of cone V$=\dfrac{1}{3}\pi {{r}^{2}}h=\dfrac{1}{3}\pi {{r}^{2}}\dfrac{\sqrt{{{S}^{2}}-2\pi S{{r}^{2}}}}{\pi r}$

V$=\dfrac{r}{3}\sqrt{{{S}^{2}}-2\pi S{{r}^{2}}}$

For maximum volume, $\dfrac{dV}{dr}=0$.

$\begin{align}

& \dfrac{1}{3}\sqrt{{{S}^{2}}-2\pi S{{r}^{2}}}+\dfrac{r}{3}\dfrac{-4\pi Sr}{2\sqrt{{{S}^{2}}-2\pi S{{r}^{2}}}}=0 \\

& \dfrac{1}{3}\sqrt{{{S}^{2}}-2\pi S{{r}^{2}}}=\dfrac{r}{3}\dfrac{2\pi Sr}{\sqrt{{{S}^{2}}-2\pi S{{r}^{2}}}} \\

& {{S}^{2}}=4\pi S{{r}^{2}} \\

\end{align}$

we know that, $S\ne 0$.

So $S=4\pi {{r}^{2}}$

Now taking (1) and substituting the $S=4\pi {{r}^{2}}$ we get,

$l=\dfrac{4\pi {{r}^{2}}-\pi {{r}^{2}}}{\pi r}=3r$

So $l=\sqrt{{{h}^{2}}+{{r}^{2}}}$

$\begin{align}

& 9{{r}^{2}}={{h}^{2}}+{{r}^{2}} \\

& h=2\sqrt{2}r \\

\end{align}$

Now, let $\theta $ be the semi-vertical angle of the cone.

$\begin{align}

& \tan \theta =\dfrac{r}{h}=\dfrac{r}{2\sqrt{2}r} \\

& \tan \theta =\dfrac{1}{2\sqrt{2}} \\

\end{align}$

So $\theta ={{\tan }^{-1}}\left( \dfrac{1}{2\sqrt{2}} \right)$

Hence proved.

Note: This question deals with maxima hence differentiation is involved. Few relations like the one between r, l & h are determined based on the geometry of the figure. For finding the condition for maximum volume, volume is differentiated wrt the radius and equated to 0. Avoid mistakes while differentiating.

Recently Updated Pages

Basicity of sulphurous acid and sulphuric acid are

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

What is the stopping potential when the metal with class 12 physics JEE_Main

The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main

Using the following information to help you answer class 12 chemistry CBSE

Why should electric field lines never cross each other class 12 physics CBSE

Trending doubts

How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Describe the poetic devices used in the poem Aunt Jennifers class 12 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Difference Between Plant Cell and Animal Cell

Change the following sentences into negative and interrogative class 10 english CBSE

State the laws of reflection of light

State and prove Bernoullis theorem class 11 physics CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE