Question

A tank contains $100\text{litres}$ of fresh water. A solution containing $1\text{gm/litre}$ of soluble lawn fertilizer runs into the tank at the rate of $1\text{litre/min}$ and the mixture is pumped out of the tank at the rate of $3\text{litres/min}$. Then, the time when the amount of fertilizer in tank is maximum, is
(correct answer + 2, wrong answer - 0.50)

• A. $28\frac{7}{9}\text{min}$
• B. $20\frac{7}{9}\text{min}$
• C. $22\frac{7}{9}\text{min}$
• D. $27\frac{7}{9}\text{min}$

Answer 1

The correct option is D $27\frac{7}{9}\text{min}$

Let the amount of salt at any time $=y\text{gm}$
At time $t\text{min}$, net volume in the tank
$=100+(1-3)\times t=100-2t$
rate of change of salt in tank = rate of salt in - rate of salt out
$\Rightarrow \frac{dy}{dt}=\frac{1\text{gm}}{l}\times \frac{1l}{\text{min}}-\frac{y}{(100-2t)}\frac{\text{gm}}{l}\times \frac{3l}{\text{min}}\Rightarrow \frac{dy}{dt}=1-\frac{3y}{100-2t}\Rightarrow \frac{dy}{dt}+\frac{3y}{100-2t}=1\cdots \left(1\right)$

$I.F.=e^{\int \frac{3}{100-2t}dt}\Rightarrow I.F.=e^{-\frac{3\ln (100-2t)}{2}}\Rightarrow I.F.=(100-2t{)}^{-3/2}$
So,
$y(100-2t{)}^{-3/2}=\int (100-2t{)}^{-3/2}dt+C$
Assuming $(100-2t{)}^{-3/2}=x^3$
$\Rightarrow 100-2t=x^{-2}\Rightarrow -2dt=-2x^{-3}dx\Rightarrow dt=x^{-3}dx$
$\Rightarrow y{x}^3=\int dx+C\Rightarrow y=x^{-2}+C{x}^{-3}\Rightarrow y=100-2t+C(100-2t{)}^{3/2}$
At $t=0,y=0$, so
$0=100+C(100{)}^{3/2}\Rightarrow C=-\frac{1}{10}$
$\Rightarrow y=(100-2t)-\frac{(100-2t{)}^{3/2}}{10}\cdots \left(2\right)$
Differentiating w.r.t. $t$, we get
$\Rightarrow \frac{dy}{dt}=-2+\frac{3(100-2t{)}^{1/2}}{10}$
For maxima or minima, $\frac{dy}{dx}=0$, we get
$\Rightarrow -2+\frac{3(100-2t{)}^{1/2}}{10}=0\Rightarrow (100-2t{)}^{1/2}=\frac{20}{3}\Rightarrow t=\frac{250}{9}\text{min}$
$\frac{d^{2}y}{d{x}^2}=\frac{3\times }{10}\times \frac{-2\left(\right(100-2t{)}^{-1/2})}{2}$

At $t=\frac{250}{9}\text{min}$
$\frac{d^{2}y}{d{x}^2}<0$

Hence, $y$ is maximum at
$t=\frac{250}{9}\text{min}=27\frac{7}{9}\text{min}$