Q:

A disease has hit a city. The percentage of the population infected t days after the disease arrives is approximated by ​p(t)=8t e^-t/11 for 0

Accepted Solution

A:
Answer:After 11 days,32.37%.Step-by-step explanation:We have been given that a percentage of the population infected t days after the disease arrives is approximated by [tex]​p(t)=8t\cdot e^{{\frac{-t}{11}}[/tex] for [tex]0<t<33[/tex].In order to find the days when percentage of infected population is maximum, we will find critical values of the given function by setting its derivative equal to zero.[tex]P'(t )=8(1)*e^{\frac{-t}{11}}+8t*(\frac{-1}{11}*e^{\frac{-t}{11}})=0[/tex][tex]8-\frac{8t}{11}=0[/tex][tex]-\frac{8t}{11}=-8[/tex][tex]-\frac{8t}{11}*\frac{-11}{8}=-8*\frac{-11}{8}[/tex][tex]t=11[/tex]This is a critical value of the function. It can be the point of minimum or point of maximum. In order to check if it is a point of maximum, we will substitute this value of t in the second derivative (Second Derivative Test).If we get the sign of second derivative as negative - then we will have a maximum at this value of t.If we get the sign of second derivative as positive - then we will have a minimum at this value of t.[tex]P''(t )=-\frac{8}{11}e^{\frac{-1}{11}}+\frac{-8}{11}e^{\frac{-t}{11}}+ 8t*\frac{1}{121}*e^{\frac{-t}{11}}[/tex][tex]P''(t )=-\frac{8}{11}e^{\frac{-1}{11}}+\frac{-8}{11}e^{\frac{-t}{11}}+ 8t*\frac{1}{121}*e^{\frac{-t}{11}}[/tex]At [tex]t=11[/tex], we have second derivative negative as:[tex]P''(11)=-\frac{8}{11e}-\frac{8}{11e}+\frac{8}{11e}[/tex][tex]P''(11)=-\frac{8}{11e}[/tex]Therefore, we do have a maximum at [tex]t=11[/tex] that is after 11 days the percentage of infected people is maximum.To find the maximum percentage, we need to substitute [tex]t=11[/tex] is the given function.[tex]​p(11)=8(11)\cdot e^{{\frac{-11}{11}}[/tex][tex]​p(11)=8(11)\cdot e^{-1}[/tex][tex]​p(11)=88\cdot \frac{1}{e}[/tex][tex]​p(11)=32.37339[/tex]Therefore, 32.37% is the maximum percent of infected population.