Answer:
76.09 pounds
Step-by-step explanation:
From the information given:
[tex]R_{in} = (\dfrac{1}{2} \ lb/gal ) *(6 \ gal/min)[/tex]
[tex]R_{in} =3 \ lb/min[/tex]
At a slower of 4 gal/min, the solution gets pumped out;
So,
[tex]R_{out} = \dfrac{4A}{100+(6-4)t }[/tex]
[tex]R_{out} = \dfrac{2A}{50+t }[/tex]
The differential of the equation is:
[tex]\dfrac{dA}{dt}+ \dfrac{2}{50+t}A=3 ---(1)[/tex]
For the linear differential equation, the integrating factor is determined as:
[tex]\int_e \dfrac{2}{50 +t}dt = e^{2 In|50+t|[/tex]
[tex]= (50+t)^2[/tex]
Multiplying [tex](50+t)^2[/tex] with the above integration factor;
[tex](50+t)^2 \dfrac{dA}{dt} + 2 (50+t) A = 3(50 +t)^2[/tex]
[tex]\dfrac{d}{dt}[ (50 +t)^2 A] = 3(50+t)^2[/tex]
[tex](50+t)^2 A = (50 +t)^3 + c[/tex]
[tex]A = (50 +t) + c (50 +t )^{-2}[/tex]
By using the given condition:
A(0) = 40
40 = 50 + c (50)⁻²
10 = c(50)⁻²
OR
c = -10 × 2500
c = -25000
A = (50 + t) - 25000(50 + t)⁻²
The number of pounds of salt in the tank after 30 min is:
A(30) = (50+30) - 25000(50 + 30)⁻²
[tex]A(30) = 80 - \dfrac{25000}{6400}[/tex]
[tex]A(30) = 76.09 \ pounds[/tex]
Thus, the number of pounds of salt in the tank after 30 min is 76.09 pounds.
