# A population is modeled by the differential equation……

\frac{dP}{dt}= 1.7\times(1-\frac{P}{6000})

For what values of P is the population increasing ?

### STEP BY STEP SOLUTION:

Let be a function

P=f(t)

We know that for a increasing function,

First derivative of a function should be strictly greater than zero:

Therefore, For a increasing function P

\frac{dP}{dt}\gt 0 ,where
\frac{dP}{dt}=1.7P(1-\frac{P}{6000})

Now,

{\frac{dP}{dt}}\gt 0


Substitute the value of

{\frac{dP}{dt}=1.7P(1-\frac{P}{6000})}

into the above expression:

{1.7P(1-\frac{P}{6000})}\gt 0

Using Distributive Property ,in the above equation

{1.7P}\times 1-\textcolor{primary}{1.7P}\times \frac{P}{6000}\gt 0

Solve the inequality for P:

Convert decimal into fraction

\frac{17P}{10}-\frac{17P}{10}\times \frac{P}{6000}\gt 0
\frac{17P}{10}-\frac{17{P^ {2}}}{60000}\gt 0
\frac{17P}{10}(1-\frac{{P}}{6000})\gt0

Multiplying both side by 10/17,

P(1-\frac{{P}}{6000})\gt0
P(\frac{{6000-P}}{6000})\gt0

Multiplying both side by -6000, we get

P(P-6000)<0

For getting above equation solution

Either

P<0 \\P-6000>0

Or,

P>0\\P-6000<0

For 1st case, i.e

P<0 \\P-6000>0

No, solution possible as P<0 and P>6000 both can’t happen simultaneously

Similarly, for 2nd case

P>0\\P-6000<0
P \in (0,6000)

Hence, values of P are:

0\lt P\lt 6000



#### Solution:

\textcolor{primary}{0\lt P\lt 6000}.