Answer
$$
N(t)=600-20\sqrt {30t}
$$
The average value of a function $N(t)$ on the interval$[0,30]$ is given by:
$$
\begin{aligned}
\text{The average value}&=\frac{1}{b-a} \int_{a}^{b}(N(t)) d t\\
&=\frac{1}{30-0} \int_{0}^{30}(600-20\sqrt {30t} )d t\\
&=200 \mathrm{cases}\\
\end{aligned}
$$
Therefore the average daily inventory is $200 \mathrm{cases}$.
Work Step by Step
$$
N(t)=600-20\sqrt {30t}
$$
The average value of a function $N(t)$ on the interval$[0,30]$ is given by:
$$
\begin{aligned}
\text{The average value}&=\frac{1}{b-a} \int_{a}^{b}(N(t)) d t\\
&=\frac{1}{30-0} \int_{0}^{30}(600-20\sqrt {30t} )d t\\
&=\left.\frac{1}{30}\left(600 t-20 \sqrt{30} \cdot \frac{2}{3} t^{3 / 2}\right)\right|_{0} ^{30} \\
&=\left.\frac{1}{30}\left(600 t-\frac{40 \sqrt{30}}{3} t^{3 / 2}\right)\right|_{0} ^{30} \\
&=\frac{1}{30}(18,000-12,000)\\
&=200 \mathrm{cases}\\
\end{aligned}
$$
Therefore the average daily inventory is $200 \mathrm{cases}$.