Question

A 200-gallon tank is currently half full of water that contains 50 pounds of salt. A solution containing 1 pounds of salt per gallon enters the tank at a rate of 6 gallons per minute, and the well-stirred mixture is withdrawn from the tank at a rate of 6 gallons per minute. How many pounds of salt are in the tank 10 minutes later? Round your answer to 2 decimal places.

I got 104.82 but it is wrong.

Please explain the answer and how to get it.

Answer 1

Why don't you explain your answer, and we can see whether your math is sound.

Did you start (with s(t) being the amount of salt present)

ds/dt = 1 - 6/100 s
s(0) = 50

?

Surely you can see that 104.82 is way off. Even with no salt draining out, only 110 lb would be present after 10 minutes!

Answer 2

my equation was y=900+C/e^(1/150)t

Answer 3

yeah, but how did you get it?
Where did the 900 and 150 come from?

With only 50 lb of salt at t=0, I don't see how it works. What is y measuring?

I started out by reasoning that with no draining,

ds/dt = 6

since since each minute 6 lbs of salt are added. OUCH! Typo, since I said 1 (because of 1 lb/gal, but I ignored the fact that 6 gal/min are added)

But, at the same time 6 of the 100 gallons drain out, taking 6/100 of the salt present with them. Hence the differential equation.