Question

Show that the right circular cone of least curved surface area and given volume has an altitude equal to $\sqrt{2}$ time the radius of the base.

Answer 1

Hint: We know that the formula of volume for a right circular cone is V = $\dfrac{1}{3}\pi {{r}^{2}}h$ It is given that the volume is constant so find the value of radius in terms of height. As it is given that the surface area is least, find the derivative of that surface area and equate it 0.

Complete step by step answer:
Let r & h be the radius & height of a cone respectively and V & S be the volume and curved surface area of the cone respectively and l is the slant height.
Given volume of cone is constant
Volume of cone(V) = $\dfrac{1}{3}$ $\pi $ ${{\left( radius \right)}^{2}}$ $\left( height \right)$
V = $\dfrac{1}{3}\pi {{r}^{2}}h$
$\Rightarrow $ h= ($\dfrac{3V}{\pi }$ )$\dfrac{1}{{{r}^{2}}}$
$\Rightarrow $ h = $\dfrac{p}{{{r}^{2}}}$ [given V is constant therefore $\dfrac{3V}{\pi }$ is also constant ]
 where p = $\dfrac{3V}{\pi }$
We know that ,
Curved surface area of a Cone(S) =$\pi rl$
Given that the cone is right circular cone ,
Therefore $l=\sqrt{{{h}^{2}}+{{r}^{2}}}$
By substituting $l$ in the equation we get
S = $\pi r\sqrt{{{h}^{2}}+{{r}^{2}}}$
$\Rightarrow $ S = $\pi r\sqrt{{{\left( \dfrac{p}{{{r}^{2}}} \right)}^{2}}+{{r}^{2}}}$
$\Rightarrow $ S = $\pi r\sqrt{\dfrac{{{p}^{2}}+{{r}^{6}}}{{{r}^{4}}}}$
$\Rightarrow $ S = $\pi \left[ \dfrac{\sqrt{{{p}^{2}}+{{r}^{6}}}}{r} \right]$
$\Rightarrow $ ${{S}^{2}}$ = ${{\pi }^{2}}\left[ \dfrac{{{p}^{2}}+{{r}^{6}}}{{{r}^{2}}} \right]$
Differentiate the above equation with respect to r we get
$\dfrac{d\left( {{S}^{2}} \right)}{dr}$ = $\dfrac{d\left( {{\pi }^{2}}\left[ \dfrac{{{p}^{2}}+{{r}^{6}}}{{{r}^{2}}} \right] \right)}{dr}$
$\Rightarrow $ $\dfrac{d\left( {{S}^{2}} \right)}{dr}$ = $\dfrac{d\left( {{\pi }^{2}}\left[ {{p}^{2}}{{r}^{-2}}+{{r}^{4}} \right] \right)}{dr}$
$\Rightarrow $ $\dfrac{d\left( {{S}^{2}} \right)}{dr}$ = ${{\pi }^{2}}\left[ -2{{p}^{2}}{{r}^{-3}}+4{{r}^{3}} \right]$
Given the cone has least curved surface area
 Which implies, $\dfrac{d\left( {{S}^{2}} \right)}{dr}$ = 0
$\Rightarrow $ ${{\pi }^{2}}\left[ -2{{p}^{2}}{{r}^{-3}}+4{{r}^{3}} \right]$ = 0
$\Rightarrow $ $4{{r}^{3}}$ = $\dfrac{2{{p}^{2}}}{{{r}^{3}}}$
$\Rightarrow $ ${{r}^{6}}$ = $\dfrac{{{p}^{2}}}{2}$
Find $\dfrac{{{\partial }^{2}}\left( {{S}^{2}} \right)}{\partial {{r}^{2}}}$ and substitute ${{r}^{6}}$ = $\dfrac{{{p}^{2}}}{2}$ in it.
$\dfrac{{{\partial }^{2}}\left( {{S}^{2}} \right)}{\partial {{r}^{2}}}$ = ${{\pi }^{2}}\left[ 6{{p}^{2}}{{r}^{-4}}+12{{r}^{2}} \right]$
Therefore by substituting ${{r}^{6}}$ = $\dfrac{{{p}^{2}}}{2}$ in the above equation we get $\dfrac{{{\partial }^{2}}\left( {{S}^{2}} \right)}{\partial {{r}^{2}}}$ is greater than 0
$\dfrac{{{\partial }^{2}}\left( {{S}^{2}} \right)}{\partial {{r}^{2}}}$ > 0
Therefore the Surface area is minimum for ${{r}^{6}}$ = $\dfrac{{{p}^{2}}}{2}$
We know that
h = $\dfrac{p}{{{r}^{2}}}$
By substituting the above value of p in the above equation we get
h = $\dfrac{\sqrt{2}{{r}^{3}}}{{{r}^{2}}}$
$\therefore $h = $\sqrt{2}r$
Therefore , we can say that the right circular cone of least curved surface area has an altitude which is equal to $\sqrt{2}$ times the radius of the base.
Hence proved.

Note:
Go through all the formulae of areas and volume of all the 3-Dimensional bodies. And read the question correctly and do not misinterpret it. It is given that the circular cone is the right circular cone which is why we can use Pythagoras theorem in this question. When they ask the question that it should be minimum or maximum you can make the derivative 0 and solve the problem.