c'=(e^-4t)(1-4t)

how do I find the maximum value of "c" and when it occurs "t"

Question

c'=(e^-4t)(1-4t)

how do I find the maximum value of "c" and when it occurs "t"

turingtest · Answer

the maximum occurs when c'=0

turingtest · Answer

max or min*

turingtest · Answer

when is e^-4t=0 
?

anonymous · Answer

when t=0?

turingtest · Answer

is it
hm... as I recall e^0=1, so that's not it...

anonymous · Answer

oh right. dang.

turingtest · Answer

what would happen if we tried to take the log of both sides?

anonymous · Answer

ummm im not sure

turingtest · Answer

remember that$$e^{-4t}=\frac1{e^{4t}}$$notice that as $t\to\infty$ we have $e^{-4t}\to0$, so t would have to be infinity for e^-4t to be zero, so it is $never$ zero

turingtest · Answer

that means we only have to solve  the possibility that the other part is zero, so solve
1-4t=0

anonymous · Answer

so t=1/4?  and then  i plug that back into the original equation?

anonymous · Answer

well i did have an original equation of c=te^-4t and im supposed to find the max value of "c" and when it occurs "t"

turingtest · Answer

okay, so the maximum value will be what you get by plugging t into that

turingtest · Answer

t=1/4 is where the maximum occures, and c(1/4) is the max value

anonymous · Answer

so to find it i put 1/4 in for t?

c=1/4 * e^-4(1/4)

turingtest · Answer

yes

anonymous · Answer

ok that makes sense. Thank you so much!

turingtest · Answer

welcome!