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). `