prove : semi verticle angle of right circular cone of given volume and least curved surface area is cot^-1 (root2) .

Let the semi-vertical angle of the cone be θ.

Volume, V=13πr2h    .....1Curved surface area, A=πrr2+h2     ......2

Since volume is constant, we can write h in terms of V from (1):-

h=3Vπr2 ------(3)

Using (3) in (2):-




Now, A is maximum or minimum when dAdr=0

So, dAdr=0  when  4π2r3+9V2-2r3=0

dAdr=0  when  4π2r3=18V2r3

dAdr=0  when  418π2r6=13πr2h2

dAdr=0  when  2r6=r4h2

dAdr=0  when  2r2=h2 

dAdr=0  when  hr=2

dAdr=0  when  cot-1θ=2

Hence Proved.

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