Question

the decreasing population, p, of owls in a national park is being monitored by ecologists and is modeled by the equation p = - 0.4t²+12t + 1,200, where t is the number of months since the ecologists started observing the owls.
a. if this model is accurate, when will the population reach its maximum?
b. what is the maximum population?
c. use the equation to determine in how many months the population of owls will disappear.
a. the population will reach its maximum in months.

Explanation:

Step1: Identify the coefficients

The population equation is $p = - 0.4t^{2}+12t + 1200$, where $a=-0.4$, $b = 12$, $c = 1200$.

Step2: Find the time for maximum population

For a quadratic function $y = ax^{2}+bx + c$, the $x$-coordinate of the vertex (which gives the time of maximum for our population - time function) is $t=-\frac{b}{2a}$. Substituting $a=-0.4$ and $b = 12$ into the formula:
$t=-\frac{12}{2\times(-0.4)}=\frac{12}{0.8}=15$ months.

Step3: Find the maximum population

Substitute $t = 15$ into the population equation $p=-0.4t^{2}+12t + 1200$.
$p=-0.4\times(15)^{2}+12\times15 + 1200$.
First, calculate $-0.4\times(15)^{2}=-0.4\times225=-90$.
Then, $12\times15 = 180$.
So, $p=-90 + 180+1200=1290$.

Step4: Find when the population disappears

Set $p = 0$, so we have the quadratic equation $-0.4t^{2}+12t + 1200=0$. Multiply through by - 10 to get $4t^{2}-120t - 12000 = 0$. Divide by 4 to simplify: $t^{2}-30t - 3000=0$.
Using the quadratic formula $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for the quadratic equation $t^{2}-30t - 3000=0$ where $a = 1$, $b=-30$, $c=-3000$.
First, calculate the discriminant $\Delta=b^{2}-4ac=(-30)^{2}-4\times1\times(-3000)=900 + 12000=12900$.
Then, $t=\frac{30\pm\sqrt{12900}}{2}=\frac{30\pm113.58}{2}$.
We get two solutions: $t_1=\frac{30 + 113.58}{2}\approx71.79$ and $t_2=\frac{30-113.58}{2}\approx - 41.79$. Since time $t\geq0$, we take $t\approx71.79\approx72$ months.

Answer:

a. 15
b. 1290
c. 72