Question

Pleasantburg has a population growth model of P(t)=at2+bt+P0 where P0 is the initial population. Suppose that the future population of Pleasantburg t years after January 1, 2012, is described by the quadratic model P(t)=0.8t2+6t+19,000. In what month and year will the population reach 34,200?

Answer 1

Answer:

The population will reach 34,200 in February of 2146.

Step-by-step explanation:

Population in t years after 2012 is given by:

$P(t) = 0.8t^{2} + 6t + 19000$

In what month and year will the population reach 34,200?

We have to find t for which P(t) = 34200. So

$P(t) = 0.8t^{2} + 6t + 19000$

$0.8t^{2} + 6t + 19000 = 34200$

$0.8t^{2} + 6t - 15200 = 0$

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$.

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})$, given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}$

$\bigtriangleup = b^{2} - 4ac$

In this question:

$0.8t^{2} + 6t - 15200 = 0$

So $a = 0.8, b = 6, c = -15200$

Then

$\bigtriangleup = 6^{2} - 4*0.8*(-15100) = 48356$

$t_{1} = \frac{-6 + \sqrt{48356}}{2*0.8} = 134.14$

$t_{2} = \frac{-6 - \sqrt{48356}}{2*0.8} = -141.64$

We only take the positive value.

134 years after 2012.

.14 of an year is 0.14*365 = 51.1. The 51st day of a year happens in February.

So the population will reach 34,200 in February of 2146.