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 At the same time, a well mixed solution of cherry cola is withdrawn at the rate of 5 liters per minute.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We are told that a well mixed solution of cherry cola is withdrawn at the rate of 5 liters per minute.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

the range of a function is the range of y-values for which it is defined, or exists. given that we are only looking at the y-values, we can already discount the first two answers as they deal with the domain of the function (range of x-values).

now, what we need to do is find the smallest and highest y-values for which the function is defined. if you'd like, you could take a ruler and start at the very bottom of the graph, find the lowest point, then move the ruler higher and higher until you reach the highest point. doing so either in your head or with an actual ruler, we can see that the lowest y-value is -5 and the highest y-value is -1.

the question now remains as to whether they are included in the range (fourth answer) or excluded (third answer). the key here is to recognise whether they are marked with an open or a closed circle. since both (-2, -5) and (2, -1) are marked with a closed circle, they are points that exist on the graph and are thus included in the range (note that if they were marked with an open circle like (4, -2) then that would signify that there is a break in the graph at that point).

thus, the fourth answer, (y | -5 ≤ y ≤ -1), is correct.

also note that usually it is good practice to scan your eye inbetween the highest and lowest point for any breaks in the graph - we do have one here at (4, -2), however -2 is still included in the range because there is another point that is included in the graph with the same y-value, ie. point (1, -2).

i hope that but if you require any further explanation don't hesitate to comment below.

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step-by-step explanation: