Question

A tank contains 100 liters of fresh water. A solution containing 1 gm/litre of soluble lawn fertilizer runs into the tank at the rate of 1 lit/min., and the mixture is pumped out of the tank at the rate of 3 litres/min. Find the time when the amount of fertilizer in the tank is maximum.

Answer 1

Let y be the amount of fertilizer in the tank at time t.

$V\left(t\right)=100gal+(1$ $gal/min-3$ $gal/min\left)\right(tmin)$

$=(100-2t)gal$

So, Rate out $=\frac{y\left(t\right)}{V\left(t\right)}$ outflow rate

$=\left(\frac{y}{100-2t}\right)3$

$=\left(\frac{3y}{100-2t}\right)\frac{lb}{min}$

Also,

Rate in $=\left(1\frac{lb}{gal}\right)\left(1\frac{gal}{min}\right)$

$=1\frac{lb}{min}$

Therefore, the rate of amount fertilizer in the tank is

$\frac{dy}{dt}$ Rate in - Rate out

$=1-\frac{3y}{100-2t}$

The above differential eqution, can be written as

$\frac{dy}{dt}+\frac{3y}{100-2t}=1$............(1)

linear differential equation of the first order

$y^{\prime }+P\left(t\right)y=f\left(t\right)$...........(2)

Based on (1) and (2), we obtain

$P\left(t\right)=\frac{3}{100-2t}$ and $f\left(t\right)=1$

The integrating factor is given by

$e\int P\left(t\right)dt$.............(3)

Use the formula (3), to get integrating factor

$e^{\int P\left(t\right)dt}=e^{\int \frac{3}{100-28}dt}=e^{-\frac{3}{2}ln|100-2t|}=e^{ln|100-2t{|}^{-3/2}}=(100-2t{)}^{-3/2}$

Thus, the solution is

$y(100-2t{)}^{-3/2}=\int (100-2t{)}^{-3/2}d+c$

$y=(100-2t{)}^{3/2}(\frac{1}{(100-2t{)}^{3/2}}+c)$

$y=100-2t+c(100-2t{)}^{3/2}$..........(4)

Use the initial condition $y\left(0\right)=0$ to get c.

$y\left(0\right)=100+c(100{)}^{3/2}$

$0=100+c(100{)}^{3/2}$

$c=-\frac{1}{10}$

Substitute the value $c=-\frac{1}{10}$ in the equation (4).

$y\left(t\right)+100-2t-\frac{1}{10}(100-2t{)}^{3/2}$

Differentiating both side with respect to t, we get,

$\frac{dy}{dt}=\frac{3\sqrt{-2t+100}}{10}-2$

Let $\frac{dy}{dt}=0$ then

$\frac{3\sqrt{-2t+100}}{10}-2=0$

$\frac{3\sqrt{-2t+100}}{10}=2$

$\sqrt{-2t+100}=\frac{20}{3}$

$-2t+100=(\frac{20}{3}{)}^2$

$=-2t=(\frac{20}{3}{)}^2-100$

$-2t=-\frac{500}{9}$

$t=\frac{500}{9}.\frac{1}{2}$

$t\approx 27.8min$

So, the maximum amount is

$y\left(27.8\right)=100-\mathrm{2.27.8}-\frac{1}{10}(100-\mathrm{2.27.8}{)}^{3/2}\approx 14.8lb$

$t\approx 27.8min$

The maximum amount is $\approx 14.8lb$