(i) The initial-value problem for the amount of salt, S(t), in the tank at time t can be formulated as follows:
dS/dt = (1/2) - (1/30)S, S(0) = 6
(ii) To solve the initial-value problem, we can use an integrating factor. Let's rewrite the equation as:
dS/dt + (1/30)S = 1/2
The integrating factor is given by the exponential of the integral of the coefficient of S:
IF = exp((1/30)t) = e^(t/30)
Multiplying both sides of the equation by the integrating factor, we have:
e^(t/30) dS/dt + (1/30)e^(t/30)S = (1/2)e^(t/30)
Applying the product rule on the left side, we get:
d/dt (e^(t/30)S) = (1/2)e^(t/30)
Integrating both sides with respect to t, we have:
e^(t/30)S = (1/2)∫e^(t/30)dt
Evaluating the integral and solving for S, we find:
S(t) = (15/2)e^(-t/30) + (1/2)
(iii) When the tank is full, it means that the amount of water entering the tank equals the amount leaving the tank. In this case, the rate at which salt water is being pumped into the tank is equal to the rate at which the well-mixed solution is leaving.
The rate at which salt water is being pumped into the tank is 2 gallons per minute, and the rate at which the well-mixed solution is leaving is 1 gallon per minute. Therefore, the tank will be full after 30 minutes.
To find the amount of salt in the tank when it is full, we substitute t = 30 into the solution obtained in step (ii):
S(30) = (15/2)e^(-30/30) + (1/2) = (15/2)e^(-1) + (1/2)
The final amount of salt in the tank when it is full can be calculated by evaluating the expression above.