Answer:
- A(0) = 2; A'(t) = 6 - 0.08A(t)
- A(∞) = 75
Explanation:
The initial value is said to be 2 pounds, so A(0) = 2.
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The influx is (1.5 #/gal)×(4 gal/min) = 6.0 #/min.
The outflow is (4 gal)/(50 gal) × A(t) = 0.08A(t).
The rate of change of A(t) is then ...
A'(t) = 6 - 0.08A(t)
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When the system reaches steady state, A'(t) = 0, so ...
0 = 6 - 0.08A(∞)
A(∞) = 6/0.08 = 75
75 pounds of salt will be in the tank after a long period.