Answer:
50 gallon
Step-by-step explanation:
We are given that
[tex]\frac{dV_{in}}{dt}=10 gal/min[/tex]
Let V(in gallon) be the present volume of tank.
[tex]\frac{dV_{out}}{dt}=0.2 V gal/min[/tex]
We have to find the smallest capacity the tank can have if the process is to continue indefinitely.
According to question
[tex]\frac{dV}{dt}[/tex]=Incoming volume-outgoing volume
[tex]\frac{dV}{dt}=10-0.2 V=-(\frac{1}{5}V-10)=-(\frac{V-50}{5})[/tex]
[tex]\frac{5dV}{V-50}=-dt[/tex]
Integrating on both sides
[tex]\int \frac{5dV}{V-50}=-\int dt[/tex]
[tex]5ln(V-50)=-t+ln C[/tex]
By using the formula
[tex]\int \frac{dx}{x}=ln x+C[/tex]
[tex]ln(V-50)=\frac{1}{5}(-t+ln C)[/tex]
[tex]ln(V-50)=-\frac{1}{5}t+\frac{1}{5} ln C[/tex]
[tex]ln(V-50)=-\frac{1}{5} t+lnC^{\frac{1}{5}}[/tex]
[tex] ln(V-50)=-\frac{1}{5}t+ln C'[/tex]
Where [tex]ln C'=ln C^{\frac{1}{5}}[/tex]
[tex] ln(V-50)-ln C'=-\frac{1}{5} t[/tex]
[tex]ln\frac{V-50}{C'}=-\frac{1}{5}[/tex]
Using the formula
[tex] ln m-ln n=ln\frac{m}{n}[/tex]
[tex]\frac{V-50}{C'}=e^{-\frac{1}{5}t}[/tex]
[tex]V-50=C'e^{-\frac{1}{5}t}[/tex]
Substitute [tex]t=\infty[/tex]
[tex]V-50=0[/tex]
[tex]V=50[/tex]
Hence, the smallest capacity the tank can have if the process is to continue indefinitely=50 gallon
