Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is `cos^(-1)(1/sqrt3)`
Solution
Let us consider the following variables:
h = height of the cone
l = slant height of the cone (given)
r = radius of the base of the cone
α = semi vertical angle of the cone
Let us assume that V' is the volume of the cone which has to be maximised.
We know that
`V' = 1/3πr^2h ...(1)`
From the figure, we have:
`l^2=r^2+h^2⇒l^2−h^2=r^2`
On substituting the value of r2 in equation (1), we get:
`V' = 1/3π(l^2−h^2)h=1/3π(l^2h−h^3) `
On differentiating with respect to h, we get:
`dV'/dh=1/3π(l^2−3h^2) ...(2)`
For maximum volume of V, let us put `(dV')/(dh)=0.`
So, from equation (2), we have:
`1/3π(l^2−3h^2)=0⇒h=l/sqrt3 (∵ h,l >0)`
Again, differentiating equation (2) with respect to h, we get:
`(d^2V'/dh^2)=1/3π(−6h)=−2πh`
`∴ ((d^2V')/(dh^2))_(h=l/sqrt3)=−2πl/sqrt3<0`
Thus, the volume of the cone is maximum at
`h=l/sqrt3`
From the figure, we have:
`cosα =h/l⇒cosα =l/lsqrt3=1/sqrt3`
`⇒α=cos^(−1)(1/sqrt3)`
∴ The semi-vertical angle is
`cos^(−1)(1/sqrt3). `
