Question5. The tank in Fig. 28 contains 80 lb of salt dissolved in 500 gal of water. The inflow per minute is 20 lb of salt dissolved in 20 gal of water. The outflow is $20 \mathrm{gal} / \mathrm{min}$ of the uniform mixture. Find the time when the salt content $y(t)$ in the tank reaches $95 \%$ of its limiting value (as $t \rightarrow$ $\infty)$.
Fig. 28. Tank in Problem 28

Studdy Solution
Calculate the time when $y(t)$ reaches 95% of its limiting value:
95% of the limiting value is:
$0.95 \times 500 = 475$
Set $y(t) = 475$ and solve for $t$:
$475 = 500 - 420e^{-\frac{t}{25}}$
$420e^{-\frac{t}{25}} = 25$
$e^{-\frac{t}{25}} = \frac{25}{420}$
Take the natural logarithm of both sides:
$-\frac{t}{25} = \ln\left(\frac{25}{420}\right)$
Solve for $t$:
$t = -25 \ln\left(\frac{25}{420}\right)$
Calculate the value:
$t \approx 75.54 \, \text{minutes}$
The time when the salt content reaches 95% of its limiting value is approximately:
$$\boxed{75.54 \, \text{minutes}}$$