A tank initially contains $50$ gal. of brine, with $30$ lbs of salt in solution. Water runs into the tank at $6$ gal./min. and the well-stirred solution runs out at $5$ gal./min. How long will it be until there are $25$ lbs of salt in the tank?
So I set up the equation and solved it and ended up with $Q=t+50-\frac{6250000000}{(t+50)^{5}}$
Since it asks for how long it will be until there are $25$ lbs of salt left, I set $Q$ (concentration equal to $25$ and tried solving for $t$.
My answer is that $t=-1.56$ so I'm assuming that means it'll take $1.56$ mins until there are $25$ lbs of salt left. However, my online homework won't accept that answer...am i approaching the problem wrong?
Let $Q$ equal the amount of salt in the container. We have that at $t=0$, $Q=30$. Additionally, we have that $Q$ satisfies $$\frac{dQ}{dt}=-\frac{Q}{W}\cdot 5$$ where $W$ is the amount of mixture. This is because we are loosing $5$ gallons of mixture that has concentration $Q/W$ every minute. We know that the amount of mixture, $W$, is given by $W=50+t$ because we start with 50 gallons and have a net gain of one gallon of mixture every minute. Thus $$\frac{dQ}{dt}=-\frac{Q}{50+t}\cdot 5$$ This is a first order linear DE (and separable DE). Thus we have the solution $$\int\frac{dQ}{Q}=-5\int\frac{dt}{50+t}$$ Giving $$\ln(Q)=-5\ln(t+50)+C$$ $$\text{and}$$ $$Q=c(t+50)^{-5}$$ Using $Q(0)=30$ we get $c=30\cdot 50^5$. Setting $Q=25$ we have $$25=30\cdot 50^5(t+50)^{-5}$$ telling us that $Q=25$ when $t=\left(\frac{25}{30\cdot 50^5}\right)^{-1/5}-50\approx 1.856864$