Answer
Verified
415.2k+ 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
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
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
Trending doubts
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE
Which type of bond is stronger ionic or covalent class 12 chemistry CBSE
What organs are located on the left side of your body class 11 biology CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
How fast is 60 miles per hour in kilometres per ho class 10 maths CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
When people say No pun intended what does that mea class 8 english CBSE