# Brine containing 3lb/gal of salt enters a tank at the

Brine containing 3lb/gal of salt enters a tank at the rate of 4gal/min and the solution well-stirred leaves at 2gal/min. The tank initially contains 20 gal of water. What is the amount of salt in the tank after 5 minutes?
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Pansdorfp6
Solution:
The salt enters the tenk at rate

$R=12$ lbs/min
Since the tenk loses liquid at a rate of 2 gal/min

after to time: (minutes). The number of gallons of Brine in the tenk
$=20+2t$
So after 5 min
$=20+2×5$
$=30$ gallons
and 2 gallons $=.3454lbs$
So amount of salt $=30×8.3454$
$=250.362$
###### Not exactly what you’re looking for?
Heather Fulton
Given that the initial amount of salt is zero( because of pure water)
Rate in 4gal/min
Rate out 2gal/min initial volume is 20 gal
Brine containing 3lb/gal of salt
The inteving rate 4gal/min
So intering concentration
$=3×4$
$=12$
Volume $=20gal$
Let the Amount of Brine is y(+)
Then rate change
${y}^{1}=rat{e}_{\mathrm{ln}}-rat{e}_{out}$
$=12-\frac{2×y}{20+\left(4-2\right)t}$
${y}^{1}=12-\frac{y}{10+t},y\left(0\right)=0$
Salving

$y×\left(10+t\right)=\int 12×\left(10+t\right)dt+c=6{\left(10+t\right)}^{2}+c$
$y\left(+\right)=6\left(10+t\right)+\frac{c}{10+t}$
By $y\left(0\right)=0⇒0=60+\frac{c}{10}⇒c=-600$
$y\left(t\right)=6\left(10+t\right)-\frac{600}{10+t}$
The amount after 5 min
$y\left(5\right)=6×\left(10+5\right)-\frac{600}{10+5}=90-\frac{600}{15}$
$y\left(5\right)=90-40=50lb$
karton