Question

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

Answer 1

After 11 days,

32.37%.

Explanation:

We have been given that a percentage of the population infected t days after the disease arrives is approximated by
`​p(t)=8t\cdot e^{{(-t)/(11)}` for
`0<t<33`.

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.

`P'(t )=8(1)*e^{(-t)/(11)}+8t*((-1)/(11)*e^{(-t)/(11)})=0`

`8-(8t)/(11)=0`

`-(8t)/(11)=-8`

`-(8t)/(11)*(-11)/(8)=-8*(-11)/(8)`

`t=11`

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.

`P''(t )=-(8)/(11)e^{(-1)/(11)}+(-8)/(11)e^{(-t)/(11)}+ 8t*(1)/(121)*e^{(-t)/(11)}`

`P''(t )=-(8)/(11)e^{(-1)/(11)}+(-8)/(11)e^{(-t)/(11)}+ 8t*(1)/(121)*e^{(-t)/(11)}`

At
`t=11`, we have second derivative negative as:

`P''(11)=-(8)/(11e)-(8)/(11e)+(8)/(11e)`

`P''(11)=-(8)/(11e)`

Therefore, we do have a maximum at
`t=11` that is after 11 days the percentage of infected people is maximum.

To find the maximum percentage, we need to substitute
`t=11` is the given function.

`​p(11)=8(11)\cdot e^{{(-11)/(11)}`

`​p(11)=8(11)\cdot e^(-1)`

`​p(11)=88\cdot (1)/(e)`

`​p(11)=32.37339`

Therefore, 32.37% is the maximum percent of infected population.