Answer
The speed the water was running out at the end of 10 minutes is $8000ga/mi$.
The average rate at which the water flows out during the first 10 min is $-10000ga/mi$.
Work Step by Step
$$Q(t)=200(30-t)^2$$
First, we need to rewrite the formula of $Q(t)$: $$Q(t)=200(30-t)^2=200(900-60t+t^2)=180000-12000t+200t^2$$
Find the rate of change of water draining $dQ/dt$:
$$\frac{dQ}{dt}=Q'(t)=\Big(180000-12000t+200t^2\Big)'=-12000+400t(ga/mi)$$
So, $$Q'(10)=-12000+400\times10=-12000+4000=-8000(ga/mi)$$
The speed the water was running out at the end of 10 minutes is $|Q'(10)|=8000ga/mi$.
We can find the average rate at which the water flows out during the first 10 minutes by dividing the difference in the amount of water in the tank at first, $Q(0)$, and after $10$ minutes, or $Q(10)$, by the amount of time.
In other words, the average rate is $$\frac{Q(10)-Q(0)}{10}=\frac{200(30-10)^2-200(30-0)^2}{10}=\frac{200\times20^2-200\times30^2}{10}$$
$$=-10000(ga/mi)$$
