Question

Prove that the semi vertical angle of the right circular cone of given volume and least curved surface area is cot ^-1(√2)

Answer 1

Solution:

Let V be the given volume of cone.

V=Volume of cone = $\frac{1}{3}\pi r^2h$

h = Height of cone = $\frac{3 V}{\pi r^2}$

C=Curved surface area of cone = $\pi r l$, where r is radius and l is slant height of cone.

C= $\pi r \sqrt{r^2 + h^2}$, as l²= r²+ h²

For Maxima and Minima, derivative of C that is curved surface area should be equal to zero.

K=C²= π²r²(r²+h²)

K =$\pi }^2 r ^4 +r^2 \times \frac{9 V^2}{(\pi)^2 r^4}=\pi }^2 r ^4 +\frac{9V^2}{ r^2 }$

Differentiating both sides with respect to r

K' = 4π²r³ + $\frac{-18 V^2}{r^3}$

Putting , K'=0

$4(\pi)^2 r^3 + \frac{-18 V^2}{r^3}=0 \\\\  r= [\frac{9V^2}{2 (\pi)^2}]^\frac{1}{6}$=$\frac{3^\frac{1}{3} V^\frac{1}{3}}{2^\frac{1}{6}(\pi)^{\frac{1}{3}}}$

h=$\frac{3 V}{\pi \times [\frac{9 V^2}{2 (\pi )^2}]^\frac{1}{3}} =(\frac{6 V}{\pi})^\frac{1}{3}$

Let A be the semi vertical angle of the cone.

Cot A =  $\frac{h}{r}$

       = $\frac{(\frac{6 V}{\pi})^\frac{1}{3}}{{\frac{3^\frac{1}{3} V^\frac{1}{3}}{2^\frac{1}{6}(\pi)^{\frac{1}{3}}}}}$

Cot A= $\sqrt{2}$

A= $Cot^{-1}{\sqrt2}$

Hence proved.

Answer 2

Answer:

Solution:

Let V be the given volume of cone.

V=Volume of cone =

h = Height of cone =

C=Curved surface area of cone = , where r is radius and l is slant height of cone.

C= , as l²= r²+ h²

For Maxima and Minima, derivative of C that is curved surface area should be equal to zero.

K=C²= π²r²(r²+h²)

K =

Differentiating both sides with respect to r

K' = 4π²r³ +

Putting , K'=0

=

h=

Let A be the semi vertical angle of the cone.

Cot A =  

       =

Cot A=

A=

Hence proved.