A large tank is partially filled with 100 gallons of fluid in which 20 pounds of salt is dissolved. Brine containing 1 2 pound of salt per gallon is pumped into the tank at a rate of 6 gal/min. The well-mixed solution is then pumped out at a slower rate of 4 gal/min. Find the number of pounds of salt in the tank after 35 minutes.

Step-by-step explanation:

From the given information:

Given that the solution is pumped at a slower rate of 4gal/min

Then:

The differential equation can be expressed as:

Integrating the linear differential equation; we have::

multiplying above integrating factor fields; we have:

A = (50 + t) + c(50 + t)²

Using the given conditions:

A(0) = 20

⇒ 20 = 50 + c (50)⁻²

-30 = c(50) ⁻²

c = -30 × 2500

c =  -75000

A = (50+t) - 75000(50 + t)⁻²

The no. of pounds of salt in the tank after 35 minutes is:

A(35) = (50 + 35) - 75000(50 + 35)⁻²

A(35) = 85 -

A(35) =69.6193 pounds

A(35) 70 pounds

Thus; the number of pounds of salt in the tank after 35 minutes is 70 pounds.

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