A 1000-liter tank initially contains a mixture of 450 liters of cola and 50 liters of cherry syrup. Cola is added at the rate of 8 liters per minute, and cherry syrup is added at the rate of 2 liters per minute. At the same time, a well mixed solution of cherry cola is withdrawn at the rate of 5 liters per minute. Let S(t) be the amount (measured in liters) of cherry syrup in the tank at time t, where t is measured in minutes.(a) Formulate an initial-value problem for the the amount of cherry syrup in the tank overtime.(b) Solve the initial-value problem using an integrating factor.(c) What percentage of the mixture is cherry syrup when the tank is full?

A) dS/dt = 2 - S/(100 + t) ; S(0) = 50

B) S(t) = (100 + t) - 5000/(100 + t)

C) 35%

Step-by-step explanation:

A) Let;

t = time in minutes

S = amount of syrup

dS/dt = rate at which syrup is flowing through tank

We are told that initially, the tank contains a mixture of 450 liters of cola and 50 liters of cherry syrup.

Thus,

At t = 0

Amount of cola = 450 litres

Amount of syrup = 50 litres

Total mixture = 450 + 50 = 500 litres

We are told that a well mixed solution of cherry cola is withdrawn at the rate of 5 liters per minute.

Thus, amount of liquid in the tank at a time "t" is; 500 + 5t

We are told that cherry syrup is added at the rate of 2 liters per minute.

Thus, rate at which syrup is flowing through tank is given as;

dS/dt = 2 - 5[S/(500 + 5t)]

Factorizing out, we have;

dS/dt = 2 - 5[S/5(100 + t)]

5 will cancel out to give;

dS/dt = 2 - S/(100 + t)

Since we have 50 liters of cherry syrup initially, the initial value problem is;

dS/dt = 2 - S/(100 + t) ; S(0) = 50

B) We have our initial value problem as;

dS/dt = 2 - S/(100 + t)

Rearranging, we have;

dS/dt + S/(100 + t) = 2

Multiplying through by 100 + t gives;

(dS/dt)•(100 + t) + S = 2(100 + t)

Integrating each term with respect to t gives;

(100 + t)S = (100 + t)² + c

Divide both sides by (100 + t) to get;

S = (100 + t) + c/(100 + t)

At t = 0;

S(0) = (100 + 0) + c/(100 + 0)

S(0) = 100 + c/100

Recall earlier in our initial value problem where we stated that S(0) = 50.

Thus;

100 + c/100 = 50

Subtract 100 from both sides to givw;

c/100 = 50 - 100

c/100 = -50

c = -50 × 100

c = -5000

Thus;

S(t) = (100 + t) - 5000/(100 + t)

C) The capacity of the tank is given as 1000 liters.

Earlier, we saw that amount of liquid in the tank at a time "t" is; 500 + 5t

Thus;

500 + 5t = 1000

5t = 1000 - 500

5t = 500

t = 500/5

t = 100 minutes

We want to find What percentage of the mixture is cherry syrup when the tank is full.

Thus we will plug in t = 100 minutes into S(t) = (100 + t) - 5000/(100 + t)

Thus;

S(100) = (100 + 100) - 5000/(100 + 100)

S(100) = 200 - 25

S(100) = 175 litres

We saw earlier that total mixture initially was 500 litres.

Thus, percentage of the mixture that is cherry syrup when the tank is full is;

175/500 × 100% = 35%

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