The largest cone and the largest cylinder inscribed in a sphere

Position of the axis. The centre $O$ of the sphere is equidistant from all points of a circle lying on the sphere, so $O$ lies on the axis of that circle (the line through its centre, perpendicular to its plane).

1. Cylinder. The two base circles have the same radius $r$, so their planes are at the same distance from $O$, on opposite sides: they are at distance $\frac h2$ from $O$, and $r^2+\left(\frac h2\right)^2=R^2$. So
$$V(h)=\pi\left(R^2-\frac{h^2}4\right)h,\qquad V'(h)=\pi\left(R^2-\frac{3h^2}4\right),$$
which vanishes at $h=\dfrac{2R}{\sqrt3}$ in $(0,2R)$, with $V'>0$ before and $V'<0$ after. Then $r^2=\frac{2R^2}3$ and
$$V_{\max}=\frac{4\pi R^3}{3\sqrt3}\approx0.577\cdot\frac{4\pi R^3}3.$$

2. Cone. The axis of the cone passes through $O$ and the apex is on the axis and on the sphere, so it is one of the two points where the axis meets the sphere, i.e. one end of a diameter. Let the base plane be at distance $h$ from the apex, $0<h<2R$. The base circle satisfies $r^2=R^2-(R-h)^2=h(2R-h)$. So
$$V(h)=\frac\pi3h^2(2R-h),\qquad V'(h)=\frac\pi3(4Rh-3h^2),$$
which vanishes at $h=\dfrac{4R}3$ (for $0<h<2R$), with $V'>0$ before and $V'<0$ after. Then
$$V_{\max}=\frac\pi3\cdot\frac{16R^2}9\cdot\frac{2R}3=\frac{32\pi R^3}{81}.$$

Comparison. The sphere has volume $\frac{4\pi R^3}3$, so the best cylinder fills $\frac1{\sqrt3}\approx57.7\%$ of it, and the best cone fills $\frac{32/81}{4/3}=\frac8{27}\approx29.6\%$.