A large tank contains 128 gallons of brine in which 10 pounds of salt is dissolved. Brine containing 5 pounds of dissolved salt per gallon runs into the tank at the rate of 10 gallons per minute. The mixture, kept uniform by stirring, runs out of the tank at the rate of 4 gallons per minute. How much salt is in the tank at the end of 13 minutes?

Question

anonymous · Answer

you mean this one?

anonymous · Answer

seems like $$\large \frac{ dQ }{ dt }=rate_{i n}-rate_{out}$$

anonymous · Answer

$\large Q(0)=\frac{10}{128} lb/gallon\
\large r_{i n}=10~gallon/min \
\large r_{out}=4~gallon/min\
 \large rate  _{i n} = 5r_{i n}~lbs\
\Large rate_{out} = \frac{Q(t)r_{out}}{128+6t}\
\Large\frac{dQ}{dt}=50-\frac{4Q(t)}{128+6t} $

anonymous · Answer

hope i did that right

anonymous · Answer

that should be the formula for the concentration of salt, if you multiply that by the amount of water again you should get the amount of salt

anonymous · Answer

Thats what I did, I am just stuck on how to separate and integrate lol.

anonymous · Answer

This isn't a separable equation. It's a linear ODE, so you can find the integrating factor, yada yada yada

anonymous · Answer

I tried that too, I just keep getting stuck. 
Thanks for the response though