Question

Between 2000 and 2020, the population of Mathville could be modeled by the function m(t)=100(cubed root(t)) where m(t) is the number of years since 2000. Between those same years, the population of Calcfield could be modeled by the function c(t)=18t.
C: Write and solve an equation to algebraically confirm where the two functions intersect. Show your work.

Answer 1

To find the intersection of the two population models, we need to set the two functions equal to each other:

1. The population of Mathville is given by:
\[ m(t) = 100 \cdot \sqrt[3]{t} \]

2. The population of Calcfield is given by:
\[ c(t) = 18t \]

To find the intersection, we set these two functions equal to each other:

\[
100 \cdot \sqrt[3]{t} = 18t
\]

Next, to eliminate the cube root, we will first divide both sides by 2 for simplification:

\[
50 \cdot \sqrt[3]{t} = 9t
\]

Next, we can solve for \( t \). We'll isolate \( \sqrt[3]{t} \):

\[
\sqrt[3]{t} = \frac{9t}{50}
\]

Now, we can eliminate the cube root by cubing both sides:

\[
t = \left( \frac{9t}{50} \right)^3
\]

Expanding the right side, we calculate:

\[
t = \frac{729t^3}{125000}
\]

Next, we multiply both sides by 125000 to eliminate the fraction:

\[
125000t = 729t^3
\]

Rearranging this equation gives us:

\[
729t^3 - 125000t = 0
\]

Now, we can factor out \( t \):

\[
t(729t^2 - 125000) = 0
\]

Setting each factor to zero gives:

1. \( t = 0 \) (This corresponds to the year 2000.)
2. \( 729t^2 - 125000 = 0 \)

Now, let's solve for \( t \) in the second equation:

\[
729t^2 = 125000
\]

Dividing both sides by 729:

\[
t^2 = \frac{125000}{729}
\]

Taking the square root of both sides:

\[
t = \sqrt{\frac{125000}{729}} = \frac{\sqrt{125000}}{\sqrt{729}} = \frac{\sqrt{125 \cdot 1000}}{27} = \frac{10 \sqrt{125}}{27}
\]

Calculating \( \sqrt{125} = 5\sqrt{5} \):

\[
t = \frac{10 \cdot 5\sqrt{5}}{27} = \frac{50\sqrt{5}}{27}
\]

Approximating \( \sqrt{5} \approx 2.236 \):

\[
t \approx \frac{50 \cdot 2.236}{27} \approx \frac{111.8}{27} \approx 4.14
\]

This indicates the intersection occurs approximately 4.14 years after 2000, or around the year **2004**.

Thus, the solution shows that both equations intersect around the year **2004**, confirming that the populations of Mathville and Calcfield are approximately equal in that year.