Question

Show that the right circular cone of least curved surface and given volume has an altitude equal to \[\sqrt 2 \] times the radius of the base.

Answer 1

Hint: We are asked to prove that the height of a right circular cone is \[\sqrt 2 \] times the radius of the base if the surface area of the cone is minimal. First, recall the formulas for volume and surface area of a right circular cone. Apply the condition for minimum of a quantity and use this to prove the given statement.

Complete step-by-step answer:
Let the radius of the base be \[r\] , the slant height be \[l\] and height of the cone be \[h\] .
Volume of a right circular cone can be written as,
 \[V = \dfrac{1}{3}\pi {({\text{radius}})^2}({\text{height}})\]
Here the volume will be
 \[V = \dfrac{1}{3}\pi {r^2}h\] (i)
Surface area of a cone is written as,
 \[S = \pi ({\text{radius)(slant height}})\]
Here, the surface area will be
 \[S = \pi rl\] (ii)
The slant height of a cone can be written as,
 \[{\left( {{\text{slant height}}} \right)^2} = {\left( {{\text{radius}}} \right)^2} + {\left( {{\text{height}}} \right)^2}\]
Here, the slant height will be,
 \[{l^2} = {r^2} + {h^2}\] (iii)
Squaring equation (ii) on both sides we get,
 \[{S^2} = {\pi ^2}{r^2}{l^2}\]
Using equation (iii) in the above equation we get,
 \[{S^2} = {\pi ^2}{r^2}({r^2} + {h^2})\]
 \[ \Rightarrow {S^2} = {\pi ^2}{r^4} + {\pi ^2}{r^2}{h^2}\] (iv)
From equation (i), we have,
 \[h = \dfrac{{3V}}{{\pi {r^2}}}\] (v)
Using equation (v) in (iv) we get,
 \[{S^2} = {\pi ^2}{r^4} + {\pi ^2}{r^2}{\left( {\dfrac{{3V}}{{\pi {r^2}}}} \right)^2}\]
 \[ \Rightarrow {S^2} = {\pi ^2}{r^4} + {\pi ^2}{r^2}\left( {\dfrac{{9V^2}}{{{\pi ^2}{r^4}}}} \right)\]
 \[ \Rightarrow {S^2} = {\pi ^2}{r^4} + \dfrac{{9V^2}}{{{r^2}}}\] (vi)
Now, we differentiate equation (vi) with respect to \[r\] and we get
 \[2S\dfrac{{dS}}{{dr}} = - 2\dfrac{{9V^2}}{{{r^3}}} + 4{\pi ^2}{r^3}\]
 \[ \Rightarrow S\dfrac{{dS}}{{dr}} = - \dfrac{{9V^2}}{{{r^3}}} + 2{\pi ^2}{r^3}\] (vii)
For minimum surface area, \[\dfrac{{dS}}{{dr}}\] should be equal to zero. Therefore, equating the L.H.S of equation (vii) to zero we get,
 \[0 = - \dfrac{{9V^2}}{{{r^3}}} + 2{\pi ^2}{r^3}\]
 \[ \Rightarrow 2{\pi ^2}{r^3} = \dfrac{{9V^2}}{{{r^3}}}\]
 \[ \Rightarrow 2{\pi ^2}{r^6} = 9{V^2}\]
Now, substituting the value of \[V\] in the above equation we get,
 \[2{\pi ^2}{r^6} = 9{\left( {\dfrac{1}{3}\pi {r^2}h} \right)^2}\]
 \[ \Rightarrow 2{\pi ^2}{r^6} = 9 \times \dfrac{1}{9}{\pi ^2}{r^4}{h^2}\]
 \[ \Rightarrow 2{r^2} = {h^2}\]
 \[ \Rightarrow h = \sqrt 2 r\]
Therefore, for the right circular cone of least surface area the height or altitude of the cone is \[\sqrt 2 \] times the radius of the base.
So, the correct answer is “ \[ h = \sqrt 2 r\]”.

Note: There are few important geometrical shapes such as cone, cylinder, cube, cuboid and sphere. You should always remember the formulas for volume and surface area of these shapes and most of the questions asked in geometry are related to these formulas. Also, remember the condition for maxima or minima of a quantity.