The paper cup of largest capacity

Geometry. The slant height is $R$. If the base radius is $r$, the base circumference $2\pi r$ equals the arc length $R\varphi$, so $\varphi=\dfrac{2\pi r}R$. The height is $h=\sqrt{R^2-r^2}$.

Volume.
$$V=\frac\pi3r^2\sqrt{R^2-r^2},\qquad0<r<R.$$

Maximisation. $V>0$, so $V$ is maximal when $V^2=\frac{\pi^2}9u^2(R^2-u)$ is, with $u=r^2\in(0,R^2)$. The derivative of $u^2(R^2-u)$ is $2uR^2-3u^2=u(2R^2-3u)$, zero at $u=\frac{2R^2}3$, with a sign change from $+$ to $-$.

Answer. $r=R\sqrt{\dfrac23}$, $h=\dfrac R{\sqrt3}$, and
$$\varphi=2\pi\sqrt{\frac23}\approx5.13\text{ rad}\approx294^\circ.$$
The maximal volume is $\dfrac{2\pi R^3}{9\sqrt3}$.

Remark. One cuts out only about $66^\circ$ of the disc.