Answer
$$8160 \pi \dfrac{in^3}{s}$$
Work Step by Step
We need to use chain rule as follows: $$\dfrac{dV}{dt}=(\dfrac{\partial V}{\partial r})(\dfrac{dr}{ dt})+(\dfrac{\partial V}{\partial h})(\dfrac{dh}{dt})$$
Now, we have: $\dfrac{\partial V}{ \partial r} \approx \dfrac{\partial }{ \partial r} (\dfrac{1}{3} \pi r^2 h)=\dfrac{2\pi r^2 h}{3}$
and
$\dfrac{\partial V}{ \partial h} \approx \dfrac{\partial }{ \partial h} (\dfrac{1}{3} \pi r^2 h)=\dfrac{\pi r^2}{3} $
and $\dfrac{dV}{dt}=(11200 \pi)(1.8)+(4800 \pi)(-2.5)= 8160 \pi \dfrac{\ in^3}{s}$
