Answer
$M=\frac{C}{2}$
Work Step by Step
Step 1. Rewriting the given formula as $R=\frac{CM^2}{2}-\frac{M^3}{3}$, we have $R'=CM-M^2$
Step 2. To find the maximum of $R'$, let its derivative $R''=0$; we have $C-2M=0$ and $M=\frac{C}{2}$
Step 3. Check $R^{(3)}=R'''=-2\lt0$; the region is concave down with a maximum.
