Answer
$$
C(x)= 0.32 x^{2}-0.00004 x^{3}
$$
$$
R(x)= 0.848 x^{2}-0.0002 x^{3}
$$
First find the profit function $P(x)$
$$
\begin{aligned}
P(x)=& R(x)-C(x) \\
=& 0.528 x^{2}-0.00018 x^{3}.
\end{aligned}
$$
The profit function is increasing on (0, 2200).
Work Step by Step
$$
C(x)= 0.32 x^{2}-0.00004 x^{3}
$$
$$
R(x)= 0.848 x^{2}-0.0002 x^{3}
$$
First find the profit function $P(x)$
$$
\begin{aligned}
P(x)=& R(x)-C(x) \\
=&\left(0.848 x^{2}-0.0002 x^{3}\right) \\
& \quad-\left(0.32 x^{2}-0.00004 x^{3}\right) \\
=& 0.528 x^{2}-0.00018 x^{3}
\end{aligned}
$$
To find any intervals where this function is increasing, set $P^{\prime}(x)=0$
$$
\begin{aligned}
P^{\prime}(x) &= 0.528 (2) x-0.00018 (3) x^{2} \\
&= 1.056 x-0.00048 x^{2} \\
&=x(1.056 -0.00048 x) \\
\Rightarrow\quad\quad\quad\quad\quad\\\
x=0 & \text { or } x=2200
\end{aligned}
$$
Now, we can use the first derivative test. Check the sign of $P^{\prime}(x)$ in the intervals:
$$
(0, 2200 ), \quad ( 2200 , 3300),
$$
(1)
Test a number in the interval $(0, 2200 ) $ say $100$:
$$
\begin{aligned}
P^{\prime}(100) &=1.056 (100)-0.00048 (100)^{2} \\
&=100.8 \\
&\gt 0
\end{aligned}
$$
we see that $P^{\prime}(x)$ is positive in that interval, so $P(x)$ is increasing on $(0, 2200)$.
(2)
Test a number in the interval $(2200, 3300 ) $ say $3000$:
$$
\begin{aligned}
P^{\prime}(13000) &=1.056 (3000)-0.00048 (3000)^{2} \\
&=-1152 \\
&\lt 0
\end{aligned}
$$
we see that $P^{\prime}(x)$ is negative in that interval, so $P(x)$ is decreasing on $(2200 , 3300)$.
So, The profit function is increasing on (0, 2200).
