Question

A disease has hit a city. The percentage of the population infected t days after the disease arrives is approximated by ​p(t)equals8 t e Superscript negative t divided by 11 for 0less than or equalstless than or equals44. After how many days is the percentage of infected people a​ maximum? What is the maximum percent of the population​ infected?



A. The percentage of infected people reaches a maximum after__ days.


B. The maximum percent of the population infected is __%.

Answer 1

Answer:

A 11 days

B. 32.37%

Step-by-step explanation:

$p(t) = 8te^(-t/11)$

percentage of infected people a maximum when p '(t) = 0

==>$p '(t) = 8(1)e^{-t/11} +8te^{-t/11}(-1/11)$

==> $p '(t) = e^{-t/11}(8 -8t/11)

now, p '(t) = 0

==> [tex]e^{-t/11}(8-8t/11) = 0

==>8 -8t/11 = 0

==> t = 88/8 = 11 days

Hence percentage of infected people reaches maximum after 11 days

maximum percent of the population infected = p(11)

==> [tex]p(11) = 8(11)e^{-11/11}$

==> p(11) = 88/e = 25.752 %

=32.37%