Answer
$$\pm 585~euros$$
Work Step by Step
Since $$R(p)=3600 p-10 p^{3}$$
Then $ R(9)= 25110~euros$ and
\begin{align*}
R'(p)&= 3600 -30 p^{2}\\
&= 1170
\end{align*}
Hence, by linear approximation
\begin{align*}
\Delta R &\approx R^{\prime}(9) \Delta p\\
&=1170 \Delta p
\end{align*}
when $p$ is raised or lowered by $0.5$, we get
\begin{align*}
\Delta R &= 1170 (\pm0.5)\\
&= \pm 585~euros
\end{align*}
