Answer
$-8000$ gallons/min,
$-10000$ gallons/min
Work Step by Step
Step 1. Given the equation $Q(t)=200(30-t)^2$, we have $\frac{dQ}{dt}=(2)200(30-t)(-1)=400t-12000$ gallons/min
Step 2. At $t=10$, we have $\frac{dQ}{dt}|_{t=10}=400(10)-12000=-8000$ gallons/min (it is negative because the amount of water is reducing.)
Step 3. $Q(0)=200(30-0)^2=180000$ gallons, $Q(10)=200(30-10)^2=80000$ gallons, and the
average rate at which the water flows out during the first 10 min is given by $\frac{\Delta Q}{\Delta t}=\frac{80000-180000}{10}=-10000$ gallons/min
