Maximum of exponential function

[Binder12345]

Homework Statement

given the formula m=n*e^(-nt) show that the maximum of this curve is at m=1/(t*e^(1)).

2. The attempt at a solution
I can show this graphically but I am curious if it is possible to do it by hand?

 

[jbriggs444]

So the "t" here is a fixed but unknown constant and n is the parameter we are adjusting, right? We have a function of one variable and are looking for its maximum value.

If a function has has a maximum, what can we say about its first derivative at that maximum?

If we want to prove that it has a maximum without graphing it, are there any theorems that we might be able to invoke?

 

[Binder12345]

If we set it equal to zero and solve that will be our maximum

 

[Charles Link]

If you write it as $y=xe^{-tx}$, and use the hint by @jbriggs444 , it should be straightforward.

 

[Binder12345]

That all makes sense my issue is how do I get the e^1 in the denominator? because isn't e^(-nt)= 0 a non real answer?

 

[haruspex]

That all makes sense my issue is how do I get the e^1 in the denominator? because isn't e^(-nt)= 0 a non real answer?

You must be differentiating the expression wrt n incorrectly. You need the product rule. If still stuck, please post all your working.

 

[haruspex]

This is strange... transiently there was another post by @Binder12345 (I think) with the right answer, then it disappeared.

 

[Binder12345]

You must be differentiating the expression wrt n incorrectly. You need the product rule. If still stuck, please post all your working.

Sorry was going to edit and accidentally deleted :


I get:
(1-nt)e^(nt)

set equal to 0 and solve:
(1-nt)e^(nt)=0 -> 1-nt=0 -> n=1/t

I'm missing my e^1 in the denominator though

 

[haruspex]

missing my e^1 in the denominator though

Then you are going wrong substituting n=1/t into the original equation.

 

[Binder12345]

Then you are going wrong substituting n=1/t into the original equation.

Yup that is exactly what I was doing wrong! :\

Thank you