Answer
The producers’ surplus is given by:
$$
\begin{aligned}
\text {Producers’ surplus }&=\int_{0}^{q_{0}}[p_{0}-S(q)] d q\\
&=1999.54
\end{aligned}
$$
Work Step by Step
Suppose the supply function for concrete is given by
$$
S(q)=100+3q^{\frac{3}{2}}+q^{\frac{5}{2}}
$$
equilibrium supply $p_0$ is when $q=9$, so we have:
$$
\begin{aligned}
p_0=S(9)&=100+3(9)^{\frac{3}{2}}+(9)^{\frac{5}{2}}\\
&=424
\end{aligned}
$$
Now, we know that:
$$
\text {Producers’ surplus }=\int_{0}^{q_{0}}[p_{0}-S(q)] d q
$$
The producers’ surplus is given by:
$$
\begin{aligned}
\text {Producers’ surplus }&=\int_{0}^{q_{0}}[p_{0}-S(q)] d q\\
&=\int_{0}^{9}\left[424-\left(100+3 q^{3 / 2}+q^{5 / 2}\right)\right] d q\\
&=\left.\left(324 q-\frac{6}{5} q^{5 / 2}-\frac{2}{7} q^{7 / 2}\right)\right|_{0} ^{9}\\
&=\left[\left(324(9)-\frac{6}{5}(9)^{5 / 2}-\frac{2}{7}(9)^{7 / 2}\right)-0\right]\\
&=2916-\frac{1458}{5}-\frac{4374}{7}\\
&=1999.54
\end{aligned}
$$
