Question

Set up an equation of a tangent to the graph of the following function.
A cone is circumscribed about a sphere of radius R. The vertex angle in the axial section of the cone is $2\alpha$. Find the area of the axial section of the cone. At what value of $\alpha$ is the area of the cone the least?

Answer 1

Area $=\frac{a^2\sin 2\alpha }{2}=$ area of three triangles $=2.\frac{1}{2}aR+\frac{1}{2}rR$

Here $r=a\sin \alpha$

Solving for $r$ and a from these and then finding the area using $r$ is above equation

$\Rightarrow a\sin 2\alpha =2R(1+\sin \alpha )$

$a=\frac{2R(1+\sin \alpha )}{\sin 2\alpha }$

Which given area $=\frac{{2R}^2{(1+\sin \alpha )}^2}{\sin 2\alpha }$