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A large mixing tank initially contains 1000 gallons of water in which 30 pounds of salt have been dissolved. Another brine solution is pumped into the tank at the rate of 4 gallons per minute, and the resulting mixture is pumped out at the same rate. The concentration of the incoming brine solution is 2 pounds of salt per gallon. If represents the amount of salt in the tank at time t, the correct differential equation for A is:__________.A.) dA/dt = 4 - .08AB.) dA/dt = 8 -.04AC.) dA/dt = 4-.04AD.) dA/dt = 2-.04AE.) dA/dt = 8-.02A

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Answer:

(B)[tex]\dfrac{dA}{dt}=8-0.004A[/tex]

Step-by-step explanation:

Volume of fluid in the tank =1000 gallons

Initial Amount of Salt in the tank, A(0)= 30 pounds

Incoming brine solution of concentration 2 pounds of salt per gallon is pumped in at a rate of 4 gallons per minute.

Rate In=(concentration of salt in inflow)(input rate of brine)

[tex]=(2\frac{lbs}{gal})( 4\frac{gal}{min})=8\frac{lbs}{min}[/tex]

The resulting mixture is pumped out at the same rate, therefore:

Rate Out =(concentration of salt in outflow)(output rate of brine)

[tex]=(\frac{A(t)}{1000})( 4\frac{gal}{min})=\frac{A}{250}[/tex]

Therefore:

The rate of change of amount of salt in the tank,

[tex]\dfrac{dA}{dt}=$Rate In-Rate out\\\dfrac{dA}{dt}=8-\dfrac{A}{250}\\\dfrac{dA}{dt}=8-0.004A[/tex]