Math Snap

STEP 1

1. The function $P(t) = 6t^2 + 110t + 3000$ models the population of the town over time.
2. $t = 0$ corresponds to the year 2000.
3. To find when the population reaches 6000, we need to solve for $t$ when $P(t) = 6000$.
4. To find the population in 2030, we need to evaluate $P(t)$ at $t = 30$.

STEP 2

1. Solve for $t$ when $P(t) = 6000$.
2. Evaluate $P(t)$ at $t = 30$.

STEP 3

Set the population function equal to 6000 to find when the population reaches this number:
$$P(t) = 6t^2 + 110t + 3000 = 6000$$

STEP 4

Subtract 6000 from both sides to set the equation to zero:
$$6t^2 + 110t + 3000 - 6000 = 0$$ $$6t^2 + 110t - 3000 = 0$$

STEP 5

Solve the quadratic equation $6t^2 + 110t - 3000 = 0$ using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 6$, $b = 110$, and $c = -3000$.
Calculate the discriminant:
$$b^2 - 4ac = 110^2 - 4 \times 6 \times (-3000)$$ $$= 12100 + 72000$$ $$= 84100$$

STEP 6

Calculate the roots using the quadratic formula:
$$t = \frac{-110 \pm \sqrt{84100}}{2 \times 6}$$ $$t = \frac{-110 \pm 290}{12}$$ Calculate the two possible values for $t$:
$$t_1 = \frac{-110 + 290}{12} = \frac{180}{12} = 15$$ $$t_2 = \frac{-110 - 290}{12} = \frac{-400}{12} = -\frac{100}{3}$$ Since $t$ represents time in years and cannot be negative, we discard $t_2$.

STEP 7

To find the population in 2030, calculate $t$ for the year 2030. Since $t = 0$ corresponds to the year 2000, $t = 30$ corresponds to the year 2030.
Evaluate $P(t)$ at $t = 30$:
$$P(30) = 6(30)^2 + 110(30) + 3000$$

Calculate $P(30)$:
$$P(30) = 6(900) + 3300 + 3000$$ $$= 5400 + 3300 + 3000$$ $$= 11700$$ The population will reach 6000 in the year $2000 + 15 = 2015$.
The population in 2030 will be 11700.