Two 100-gallon tanks initially contain pure water. Brine containing 2 lb of salt per gallon enters the first tank at the rate of 1 gal/min, and then the mixed solution drains into the second tank at the same rate. There, it is again thoroughly mixed and drains at the same rate, 1 gal/min.
Let S(t)=S(1)-S(2). Intuitively, S(t)>=0 for all t. At what time is the excess S(t) maximized? What is the maximum excess?
To find the maximum excess of S(t) and the time at which it occurs, we need to first express S(t) in terms of time.
Let's analyze the situation step by step:
1. Let's denote the amount of salt in Tank 1 at time t as S1(t) and the amount of salt in Tank 2 at time t as S2(t). Since both tanks initially contain pure water, S1(0) = S2(0) = 0.
2. At time t, the amount of salt entering Tank 1 is 2 lb/gal * 1 gal/min = 2 lb/min. Since Tanks 1 and 2 both have a flow rate of 1 gal/min, the amount of salt leaving Tank 1 and entering Tank 2 at time t is S1(t)/100 lb/min.
3. The rate of change of the amount of salt in Tank 1, dS1/dt, is given by the difference between the amount of salt entering and leaving Tank 1:
dS1/dt = 2 lb/min - (S1(t)/100 lb/min)
4. Similarly, the rate of change of the amount of salt in Tank 2, dS2/dt, is given by the difference between the amount of salt entering and leaving Tank 2:
dS2/dt = (S1(t)/100 lb/min) - (S2(t)/100 lb/min)
5. The excess amount of salt, S(t), is defined as the difference between the amounts of salt in Tank 1 and Tank 2:
S(t) = S1(t) - S2(t)
Now, let's find the maximum excess of S(t) and the time at which it occurs:
1. To find the critical points (where the maximum occurs), we set dS(t)/dt = 0:
dS(t)/dt = d(S1(t) - S2(t))/dt = dS1(t)/dt - dS2(t)/dt = 0
Substitute the expressions for dS1(t)/dt and dS2(t)/dt:
2 lb/min - (S1(t)/100 lb/min) = (S1(t)/100 lb/min) - (S2(t)/100 lb/min)
Rearrange and combine like terms:
2 lb/min = 2(S1(t) - S2(t))/100 lb/min
Simplify:
200 lb/min = 2(S1(t) - S2(t))
Divide by 2:
100 lb/min = S1(t) - S2(t)
2. Now we have an equation relating the differences in salt quantities in the two tanks. Let's substitute S(t) = S1(t) - S2(t) into the equation:
100 lb/min = S(t)
This equation indicates that the maximum excess of salt, S(t), is 100 lb/min.
3. To find the time at which the maximum occurs, we need to solve for t in terms of S(t). We can do this by using the equations for dS1(t)/dt and dS2(t)/dt:
dS1(t)/dt = 2 lb/min - (S1(t)/100 lb/min) = 0
dS2(t)/dt = (S1(t)/100 lb/min) - (S2(t)/100 lb/min) = 0
Solving these equations simultaneously will give us the time t0 at which S(t) is maximized.
4. Substitute S(t) = 100 lb/min into dS1(t)/dt:
2 lb/min - (S1(t)/100 lb/min) = 0
Rearrange and solve for S1(t):
S1(t)/100 lb/min = 2 lb/min
S1(t) = 200 lb/min
Now substitute S1(t) = 200 lb/min into dS2(t)/dt:
(S1(t)/100 lb/min) - (S2(t)/100 lb/min) = 0
Substitute S1(t) = 200 lb/min:
(200 lb/min)/100 lb/min - (S2(t)/100 lb/min) = 0
Simplify:
2 - (S2(t)/100 lb/min) = 0
Solve for S2(t):
S2(t)/100 lb/min = 2 lb/min
S2(t) = 200 lb/min
Therefore, at time t0 when S(t) is maximized, the amount of salt in Tank 1 is 200 lb/min and the amount of salt in Tank 2 is 200 lb/min.
5. Finally, to find the time at which the maximum excess occurs, we need to determine how long it takes for 200 lb of salt to flow from Tank 1 to Tank 2 and accumulate there. Since the flow rate from Tank 1 to Tank 2 is 1 gal/min, it will take 200 min for 200 lb to flow to Tank 2.
Therefore, the maximum excess S(t) is 100 lb/min, and it occurs at t0 = 200 min.