-9.3.15 A tank contains 200 gal of fresh water

[karush]

$\tiny{9.3.15}$
4000
A tank contains 200 gal of fresh water. A solution containing 2 lb/gal of soluble lawn fertilizer runs into the tank at the rate of 1 gal/min, and the mixture is pumped out of the tank at the rate of 5 gal/min. Find the maximum amount of fertilizer in the tank and the time required to reach the maximum.

 

[karush]

this very mymathlab problem was also posted on mhf but was ?

and its due today ... but having got all the preveous problems right got very baffled on this one
the sanple problem was of little help.

I thought it was just a simple related rates but not.

 

[HallsofIvy]

$\tiny{9.3.15}$
A tank contains 200 gal of fresh water. A solution containing 2 lb/gal of soluble lawn fertilizer runs into the tank at the rate of 1 gal/min, and the mixture is pumped out of the tank at the rate of 5 gal/min. Find the maximum amount of fertilizer in the tank and the time required to reach the maximum.

Let X(t) bet the amount of fertilizer, in lbs, in the tank at time t, in min, with t= 0 when they begin pumping the solution in. Initially the tank contains no fertilizer so X(0)= 0. The tank originally contained 200 gals, then 1 gal per minute of liquid comes in and 5 gal per minute goes out, so there is a net loss of 1- 5= -4 gals per minute. The amount of liquid in the tank at time t is 200- 4t gal.

The concentration of fertilizer, in lbs per gallon, in the tank at time t is the amount of fertilizer in the tank, in lbs., at that time divided by the amount of liquid, in gallons, $$\frac{X(t)}{200- 4t}$$. Since there are 5 gallons per minute going out, the amount of fertilizer going out is 5 gallons/minute times $$\frac{X(t)}{200- 4t}$$ pounds/gallon= $$\frac{1000- 20t}{200- 4t}$$ pounds per minute.

So the amount of fertilizer in the tank changes because there is, at each time, t, 2 pounds/min of fertilizer going in and $$\frac{1000- 20t}{200- 4t}$$ pounds per minute going out. So rate at which the amount of fertilizer is changing is $$\frac{dX}{dt}= 2- \frac{1000- 20t}{200- 4t}= \frac{400- 8t- 1000+ 20t}{200- 4t}= \frac{12t- 600}{200- 4t}= \frac{3t- 150}{50- t}$$. Write that as $$dX= \frac{3t- 150}{50- t}dt$$ and integrate both sides. There will be, of course, a "constant of integration". Use the initial condition X(0)= 0 to determine the value of that constant.

 

[karush]

Write that as $\displaystyle dX= \frac{3t- 150}{50- t}dt$ and integrate both sides. There will be, of course, a "constant of integration". Use the initial condition X(0)= 0 to determine the value of that constant.

do you mean like this

\begin{align*}\displaystyle
\int_a^b dX&= \int_a^b \frac{3t- 150}{50- t} \, dt \\
x&=\left[3t\right]_a^b
\end{align*}

looks

not sure where $x(0)=0$ goes

 

[MarkFL]

I would let $F(t)$ be the amount of fertilizer in the tank at time $t$. From the problem statement, we get the following IVP:

\(\displaystyle \d{F}{t}=2-\frac{F}{4(50-t)}\) where $F(0)=0$

Write the ODE in linear form:

\(\displaystyle \d{F}{t}+\frac{F}{4(50-t)}=2\)

Now, we need to compute the integrating factor:

\(\displaystyle \mu(t)=\exp\left(\int \frac{1}{4(50-t)}\right)=\left(4(50-t)\right)^{-\frac{1}{4}}\)

Multiplying through by $\mu(t)$, we obtain:

\(\displaystyle \left(4(50-t)\right)^{-\frac{1}{4}}\d{F}{t}+\left(4(50-t)\right)^{-\frac{5}{4}}F=2\left(4(50-t)\right)^{-\frac{1}{4}}\)

Next, we observe that the LHS may we written as:

\(\displaystyle \frac{d}{dt}\left(\left(4(50-t)\right)^{-\frac{1}{4}}F\right)=2\left(4(50-t)\right)^{-\frac{1}{4}}\)

Okay, what do you get when you integrate through w.r.t $t$?

 

[karush]

$\displaystyle \frac{d}{dt}\left(\left(4(50-t)\right)^{-\frac{1}{4}}F\right)=2\left(4(50-t)\right)^{-\frac{1}{4}}$
integrate both sides
$\displaystyle (4(50-t))^{-1/4}F=\frac{4\sqrt{2}}{3}(50-t)^{3/4}$
$F=\frac{8}{3}(50-t)$the answer was 29.5 min and 32.8 lb

 

[MarkFL]

I made a mistake in my above post...the IVP should be:

\(\displaystyle \displaystyle \d{F}{t}=2-\frac{5F}{4(50-t)}\) where $F(0)=0$.

So, this leads to:

\(\displaystyle \mu(t)=\left(4(50-t)\right)^{-\frac{5}{4}}\)

\(\displaystyle \left(4(50-t)\right)^{-\frac{5}{4}}\d{F}{t}+5\left(4(50-t)\right)^{-\frac{9}{4}}F=2\left(4(50-t)\right)^{-\frac{5}{4}}\)

\(\displaystyle \frac{d}{dt}\left(\left(4(50-t)\right)^{-\frac{5}{4}}F\right)=2\left(4(50-t)\right)^{-\frac{5}{4}}\)

Integrate:

\(\displaystyle \left(4(50-t)\right)^{-\frac{5}{4}}F=2\left(4(50-t)\right)^{-\frac{1}{4}}+C\)

\(\displaystyle F(t)=2\left(4(50-t)\right)+C\left(4(50-t)\right)^{\frac{5}{4}}\)

We know:

\(\displaystyle F(0)=2\left(4(50)\right)+C\left(4(50)\right)^{\frac{5}{4}}=0\)

Hence:

\(\displaystyle C=-2(200)^{-\frac{1}{4}}\)

And so:

\(\displaystyle F(t)=2\left(4(50-t)\right)-2(200)^{-\frac{1}{4}}\left(4(50-t)\right)^{\frac{5}{4}}\)

I will leave it to you to show that the maximum occurs at:

\(\displaystyle t=\frac{738}{25}=29.52\)

 

[karush]

again my great thanks.
ill probably just try another similar problem.

really appreciate all the steps you showed.