Max Cone Volume

We know that the radius of the circular paper $R$ is a constant and it is also equal to the slant high of the cone.

$R^{2}=r^{2}+h^{2}$

$r^{2}=R^{2}-h^{2}$ ------ 1

The volume of the cone can be calculated by the formula below.

$V = \frac{1}{3}\pi r^{2}h$ ------- from 1 we replace $r^{2}$ by $R^{2}-h^{2}$

$V = \frac{1}{3}\pi (R^{2}-h^{2})*h = \frac{\pi}{3} (R^{2}*h-h^{3})$ , then to know the Max volume we should find $\frac{dV}{dh}$ and make it equal to zero

$\frac{dV}{dh} = \frac{\pi}{3} (R^{2}-3h^{2})$ , and when $\frac{dV}{dh}=0$ then: $h^{2}=\frac{R^{2}}{3}$ , back to 1 again $r^{2}=R^{2}-\frac{R^{2}}{3}$ , $r = \sqrt{\frac{2}{3}}*R$

Angle $\theta$ can be found from formula below.

$\theta = \frac{2 \pi r}{2 \pi R}* 360$

$\theta = \frac{2 \pi \sqrt{\frac{2}{3}}*R}{2 \pi R}* 360 = \sqrt{\frac{2}{3}} * 360$

$\theta = 293.9 \approx \boxed{294}$

Vol of cone , V = 1/3 πr^2h. We can make relation of r with θ, r = (θ/2π) R, and h = √(R^2-r^2 ), V = 1/3 πr^2√(R^2-r^2 ) , dV/dθ = dV/dr . dr/dθ = (R/2π) (2πr/3)(R^2-3/2 r^2)/ √(R^2-r^2 ) = 0,then r = R√(2/3) , then θ = 294°