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Suppose a tank with a total capacity of 60 Gallons is currently only half full of a solution of water with 2% bleach concentration. At time t=0 water with a bleach concentration of 7% is pumped in at a rate of 2 gallons per minute. Water is drained from the tank at a rate of 1 gallon per minute. When the tank is full, everything shuts off. What is the amount of bleach in the tank at this time, and what is the concentration?

I don't need it worked out I just want to know if I set this up right. let me know what i need to add or change anything.

dA/dt = (2)(.07)-(1)(A/30+t)

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Looks good to me. If you have to solve it then you will need to write down an initial condition too. –  David Feb 5 at 5:58
    
yeah in this case that would be A(0)=.6 right? –  joe Feb 5 at 6:00
    
Yes, that's right. –  David Feb 5 at 6:01
    
Now when I solve for t when the tank is 60 gallons i get 827.89 mins which seems to much for me i would think it takes 30 mins to fill up the rest of the tank. 2 gallons/min (pumped in) and 1 gallon/min (pumped out) which means getting 1 gallon/min so 30 mins –  joe Feb 5 at 6:04
    
Not sure what you mean "solve for $t$", the time taken is $30$ minutes, exactly as you said in the middle of your comment. And the question asks you to find $A$ when the tank is full, not $t$. –  David Feb 5 at 6:22

1 Answer 1

How much bleach is lost in dt?

-C/Vdt

Where C is the current amount of bleach and V is the current volume.

How much bleach is added in dt?

0.07*2*dt

So:

dC/dt = 0.07*2-C/V

And V = 30+t

Describes the amount of bleach in the tank, with the appropriate initial conditions

And you just need it at the time when the tank is full.

eta: I guess your answer is right up to missing parentheses

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