A tank contains $200 gal$ brine at the beginning and $15 lb$ salt in it! $t=0$, brine $4 \frac{lb}{gal}$ flows in with a speed of $3.5\frac{gal}{min}$. The out flow of the tank is $4\frac{gal}{min}$
$$\frac{dx}{dt}=14-\frac{4x}{200-.5t}$$
Solving...
$$\frac{x}{(400-t)^8}=2(400-t)^{-7}+c$$
$$c=\frac{-785}{400^8}$$
We want to determine how much salt is left in the tank when it has only 50 gal of water!
My attempt
$200-.5t=50 \iff t=300$
By setting t-300 into the equation to solve for x-the salt left.
Is this the right way to do it? Or something is wrong!
Your method seems fine. Well done!