Question

Two cities have insect problems. The insects in City A increase exponentially according to the equation y = 120,000e0.04t. The insects in City B increase exponentially according to the equation y = 90,000e0.08t. In 15 years, which city will have more insects, and by approximately how much?

Answer 1

Given:

     City A : $y=120,000e^0^.^0^4^t$

     City B : $y= 90,000e^0^.^0^8^t$

       $y = 90,000e^0^.^0^8^t = 90,000e^0^.^0^4^t .e^2$

          $= e^2 90000e^0^.^0^4^t$

       $y = 120000e^0^.^0^4^t = \frac{4}{3} 90000e^0^.^0^4^t$

 Hence, $e^2 > \frac{4}{3},$ then city B is increasing faster.

Answer 2

Given: City A population increased exponentially $y_A = 120,000e^{0.04t$ and City B population increased exponentially $y_B = 90,000e^{0.08t$.

Ratio of the population of city A and city B is,

$\frac{y_A}{y_B} =\frac{120,000e^{0.04t}}{90,000e^{0.08t}}$

$\frac{y_A}{y_B} =\frac{4e^{0.04t-0.08t}}{3}$

$\frac{y_A}{y_B} =\frac{4e^{-0.04t}}{3}$

After 15 years

$\frac{y_A}{y_B} =\frac{4e^{-0.04\times 15}}{3}$

$\frac{y_A}{y_B} =\frac{4e^{-0.6}}{3}$

$\frac{y_A}{y_B} =\frac{4\times 0.55}{3}$

$\frac{y_A}{y_B} =\frac{2.2}{3}$

$\frac{y_A}{y_B} =0.73$

The ratio of the population of city A and city B is less than 1. Hence city B has more population compared with city A after 15 years.