\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 0Using Distributive Property ,in the above equation
{1.7P}\times 1-\textcolor{primary}{1.7P}\times \frac{P}{6000}\gt 0Solve 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})\gt0Multiplying both side by 10/17,
P(1-\frac{{P}}{6000})\gt0P(\frac{{6000-P}}{6000})\gt0Multiplying 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:
The required answer is
\textcolor{primary}{0\lt P\lt 6000}. Similar Post:
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