Answer
$t=20(\sqrt[3] 2 -1)$ minutes
Work Step by Step
To find the amount of the chemical in the tank, we obtain:
$$\frac{dA}{dt}=c_1r_1-c_2r_2$$
We are given:
$c_1=1$
$r_1=3$
$r_2=2$
As $V(t)=r_1 \Delta t-r_2 \Delta \rightarrow \frac{dV}{dt}=1$
Integrate:
$ \int dV=\int dt=t+C$
Since $V=20 \rightarrow V=20+t $
Putting the values:
$$\frac{dA}{dt}=3\times1-2\frac{A}{V}$$
$$\frac{dA}{dt}+\frac{A}{20+t}=3$$
Multiply both sides by $(20+t)^2$
$$\frac{dA}{dt}(20+t)^2+2A(20+t)=3(20+t)^2$$
where $c$ is a constant of integration.
$$\frac{dA}{dt}[(20+t)^2A]=3(20+t)^2$$
Solve for $A$:
$$A=\frac{(20+t)^3+C}{(20+t)^2}$$
Since $A(0)=0 $
Find $C$:
$C=-20^3$
The concentration is given by:
$$c_2=\frac{A}{20+t}=\frac{20+t)^3-20^3}{(20+t)^3}$$
Putting the values:
$$\frac{20+t)^3-20^3}{(20+t)^3}=\frac{1}{2}$$
$$t=20(\sqrt[3] 2 -1)$$ minutes
