A tank contains 40 gallons of a solution composed of water and bleach. At the start there are 4 gallons of bleach in the tank. A second solution containing half water and half bleach is added tothe tank at the rate of 4 gallons per minute. The solution is kept throughly mixed and drains from the tank at the rate of 4 gallons per minute. (a) How much blcach is in the tank after t minutes?
(b) Will there be at least 14 gallons of bleach in the tank after 10 minutes?
(c) How much bleach will there be in the tank after a very long time has passed?

a) x* e∧ t/10  =  20 *e∧ t/10 + 2

b) x  =  14,2 gall. In 10 min. we get 14,2 gallons of bleach in the tank

c) After a very long time we will find 20 gallons of bleach in the tank

Step-by-step explanation:

The variation of bleach in water in the tank is Δ(x)t:

Δ(x)t  = [Amount of x added to the tank - Amount of x drain out of the tank]*Δt

Now the amount of x added to the tank is:

Rate of adding* the concentration

Rate of adding =  4 gallons per minute

Concentration  half and half ( water + bleach)  = 0,5

Rate of draining =  4 gallons per minute

Concentration of draining: Unknown but can be expressed as:

x(t)/40. According to this

Δ(x)t  =  [4 * 0,5  -  4 * x(t)/40]*Δt

Dividing by Δt on both sides of the equation and tacking limits we get

dx/ dt    =  2  - x / 10

dx/dt + x/10 =2

Multyiplying by e∧ t/10 on both sides

e∧ t/10* [ dx/dt  +  x/10 ]  =  2 *e∧ t/10

To solve it, integrating the first member is

x* e∧ t/10  =  20 *e∧ t/10 + C

for initial condition t = 0

4    =  20 + C

C = -16

a) x* e∧ t/10  =  20 *e∧ t/10 - 16

b) in 10 minutes

x* e∧ t/10  =  20 *e∧ t/10 - 16

x *e  =  20*e  - 16

x   =   20  -  16/e

x   =  20  -  16/2,718

x   =   20 -  5,88

x  =  14,2  gallons.

c) After a very long time we will find 20 gallons of bleach in the tank

.632

step-by-step explanation:

6 * 1/10 = 3/5

3*0.01 = 0.03

3/5 + 0.03 = .63

2*0.002 = .002

.63 + 0.002 = .632

b