348. Given a circle of radius $R$ with center at point $O$, $AB$ is its diameter, point $M$ is on the radius $OA$, and $|AM|:|MO|=k$. A chord $CD$ is drawn through point $M$. What is the maximum value of the area of quadrilateral $ACBD$?
348. We have: $S_{A C B D}=\frac{|A B|}{|M O|} S_{O C D}=2(k+1) S_{O C D}$. Therefore, $S_{A C B D}$ will be the largest when the area of triangle $O C D$ is the largest. But triangle $O C D$ is isosceles with the lateral side equal to $R$; hence, its area is maximized when the sine of the angle at vertex $O$ reaches its maximum. Let this angle be denoted by $\varphi$. Clearly, $\varphi_{0} \leqslant \varphi \leqslant \pi / 2$, then the value of $\varphi_{1}=\varphi_{0}$. Answer: if $k \leqslant \sqrt{2}-1$, then $S_{\max }=(k+1) R^{2}$; if $k > \sqrt{2}-1$, then $S_{\max }=2 R^{2} \sqrt{k(k+2)} /(k+1)$.