Question

A tank initially holds 80 gal of a brine solution containing 1/8 lb of salt per gallon. at t = 0, another brine solution containing 1 lb of salt per gallon is poured into the tank at the rate of 4 gal/min, while the well-stirred mixture leaves the tank at the rate of 8 gal/min. find the amount of salt in the tank when the tank contains exactly 40 gal of solution.

Answer 1

Let $A(t)$ be the amount of salt (in lbs) in the tank at time $t$. Then

$\dfrac{\mathrm dA(t)}{\mathrm dt}=\dfrac{1\text{ lb}}{1\text{ gal}}\dfrac{4\text{ gal}}{1\text{ min}}-\dfrac{A(t)\text{ lbs}}{80+(4-8)t)\text{ gal}}\dfrac{8\text{ gal}}{1\text{ min}}$
$\dfrac{\mathrm dA(t)}{\mathrm dt}=4-\dfrac{2A(t)}{20-t}$
$\dfrac{\mathrm dA(t)}{\mathrm dt}+\dfrac{2A(t)}{20-t}=4$
$\dfrac1{(20-t)^2}\dfrac{\mathrm dA(t)}{\mathrm dt}+\dfrac{2A(t)}{(20-t)^3}=\dfrac4{(20-t)^2}$
$\dfrac{\mathrm d}{\mathrm dt}\left[\dfrac{A(t)}{(20-t)^2}\right]=\dfrac4{(20-t)^2}$
$\dfrac{A(t)}{(20-t)^2}=\dfrac4{20-t}+C$
$A(t)=4(20-t)+C(20-t)^2$

Given that $A(0)=\dfrac{1\text{ lb}}{8\text{ gal}}\times(80\text{ gal})=10\text{ lbs}$, we have

$10=4(20-0)+C(20-0)^2\implies C=-\dfrac7{40}$

so that the amount of salt in the tank is given by

$A(t)=4(20-t)-\dfrac7{40}(20-t)^2$
$A(t)=10+3t-\dfrac7{40}t^2$

which is valid for $0\le t\le20$, since the tank will be empty when $80+(4-8)t=0$.

The tank will contain 40 gal of solution when $80+(4-8)t=40\implies t=10$, at which point the amount of salt in the tank would be

$A(10)=10+3(10)-\dfrac7{40}(10)^2=\dfrac{45}2=22.5\text{ lbs}$