4. (11th "Hope Cup" Invitational Competition Question) The lateral surface development of a cone is a sector with a radius of 1 and a central angle of $\frac{3}{2} \pi$. Then the maximum value of the area of the section passing through the vertex of the cone is ( ).
A. $\frac{\sqrt{3}}{4}$
B. $\frac{1}{2}$
C. $\frac{\sqrt{3}}{2}$
D. $\frac{3 \sqrt{7}}{16}$
4. B. Reason: The slant height $l=1$, base radius $r=\frac{3}{4}$, and the vertex angle of the axial section is $2 \arcsin \frac{3}{4}>\frac{\pi}{2}$, so the maximum area of the section is
$$
\frac{1}{2} \text {. }
$$