A rumour spreads through a population in such a way that t hours after the rumour starts....
...the percent of people involved in passing it on is given by P(t)=100(e^-t - e^-4t). What is the highest percent of people involved in spreading the rumour within the first 3 hours? When does this occur?
`P(t) = 100(e^-t-e^(-4t))`
To find the maximum percentage we first find the critical points when `(dP(t))/(dt) =0`
`(dP(t))/(dt) = 100(-e^-t+4e^(-4t)) = 0`
We divide both sides by 100 and move the `e^-t` to the other side and get
` e^(-t) = 4e^(-4t)`
, dividing both sides by `e^(-4t)` we get
`e^(3t) = 4` taking the ln of both sides gives us
` 3t = ln(4)`
so divide both sides by 3 to get our critial point
`t = ln(4)/3 ~~ 0.462` hours.` ` ` `
`P(0) = 100(e^0 - e^0) = 0` , and `P(3) = 100(e^(-3)-e^(-12) = 4.978`
and `P(ln(4)/3)=43.359` . So obviously at `t ~~ 0.462` hours when 43.359% of people are spreading the rumor is our answer.