89. For $\triangle A B C$, the circumradius is $R$, the perimeter is $P$, and the area is $K$. Determine the maximum value of $\frac{K P}{R^{3}}$.
89. By the Law of Sines, $P=a+b+c=2 R(\sin A+\sin B+\sin C), K=\frac{a b \sin C}{2}=2 R^{2} \sin A \sin B \sin C$.

Therefore, by the AM-GM inequality,
$$\begin{aligned}
\frac{K P}{R^{3}}= & 4 \sin A \sin B \sin C(\sin A+\sin B+\sin C) \leqslant \\
& \frac{4}{27}(\sin A+\sin B+\sin C)^{4}
\end{aligned}$$

By Jensen's inequality, $\sin A+\sin B+\sin C \leqslant 3 \sin \frac{A+B+C}{3}=\frac{3 \sqrt{3}}{2}$, so $\frac{K P}{R^{3}} \leqslant \frac{27}{4}$. Equality holds if and only if $\triangle A B C$ is an equilateral triangle.