Question

The population of a culture of Pseudomonas aeruginosa bacteria is given by P= -1600t^2 + 58000t + 30000, where to is time in hours since the culture was started. Determine the time(s) at which the population was 100000. Round to the nearest hour.

Answer 1

Answer:

The two results were 35 hours and 1.25 hours, approximately one hour fifteen minutes. Although both times meet the condition, the response of choosing one or the other time will depend on the growth behavior of P. aeruginosa

Explanation:

To determine the time(s) in which the population was 100000, you must solve the quadratic equation, clearing t, as follows:

P=-1600t^2+58000t+30000=100000

Sorting the equation and equalizing zero:

1600t^2-58000t+70000=0

 a              b                c

applying the quadratic equation is:

$t=\frac{-b+-\sqrt{b^{2}-4ac } }{2a}$

This equation has two solutions:

$t1=\frac{-b+\sqrt{b^{2}-4ac } }{2a}$

$t2=\frac{-b-\sqrt{b^{2}-4ac } }{2a}$

replacing the values ​​in both equations, we have:

$t1=\frac{-(-58000)+\sqrt{(-58000)^{2}-4(1600)(70000) } }{2(1600)}= 35$

$t2=\frac{-(-58000)-\sqrt{(-58000)^{2}-4(1600)(70000) } }{2(1600)}=1.25$

The two results were 35 hours and 1.25 hours, approximately one hour fifteen minutes. Although both times meet the condition, the response of choosing one or the other time will depend on the growth behavior of P. aeruginosa