Thread: Quick question about Optimization

"A rumour spreads through a population in such a way that t hours after its beginning, the percentage of people involved in passing the rumour is given by P(t) = 100(e^(-1) - e^(-4t)). What is the highest percentage of people involved in spreading the rumour within the first 3 h? When does this occur?"

Given the equation, P(t) = 100(e^(-1) - e^(-4t)), this is what I've done so far:

P(t) = 100(e^(-1) - e^(-4t))
P(t) = (100/e) - (100e^(-4t))
P'(t) = -100e^(-4t)(-4)
P'(t) = 400e^(-4t)

Where do I go from this point? If I make the equation equal to zero, I won't be able to solve for the time.

Thanks.

Originally Posted by Jeavus

"A rumour spreads through a population in such a way that t hours after its beginning, the percentage of people involved in passing the rumour is given by P(t) = 100(e^(-1) - e^(-4t)). What is the highest percentage of people involved in spreading the rumour within the first 3 h? When does this occur?"

Given the equation, P(t) = 100(e^(-1) - e^(-4t)), this is what I've done so far:

P(t) = 100(e^(-1) - e^(-4t))
P(t) = (100/e) - (100e^(-4t))
P'(t) = -100e^(-4t)(-4)
P'(t) = 400e^(-4t)

Where do I go from this point? If I make the equation equal to zero, I won't be able to solve for the time.

Thanks.

first off. your derivative is wrong. you have to distribute the 100 to both terms. secondly, you are looking for the absolute max i suppose. if there is no extrema to test, check the end points. this function is clearly increasing, so it should be at t = 3

Originally Posted by Jhevon

first off. your derivative is wrong. you have to distribute the 100 to both terms. secondly, you are looking for the absolute max i suppose. if there is no extrema to test, check the end points. this function is clearly increasing, so it should be at t = 3

I did distribute 100 to both terms.

The book gives these answers: 47.25% when t = 0.46h.

If that helps at all.

Originally Posted by Jeavus

I did distribute 100 to both terms.

The book gives these answers: 47.25% when t = 0.46h.

If that helps at all.

How sly you are, Jeavus, you changed it just before i clicked the "quote" button. I know what i saw, i am not crazy

anyway, that cannot be right. The graph does not even reach 40 in the first 3 hours. see the graph below: the red line is the graph, the green line is where t = 3. Note that the graph does not pass 40

check to see that you asked the right question

Haha, you're correct in the edit. :P

Anyway, I am checking the right question, I guess the book is wrong for this one.