Originally Posted by

**drumist** I'm not sure why you are taking a limit. There is no limit involved in the calculation.

You're correct that I was looking for relative maximums because that is usually the method used to solve for the absolute maximum.

Since we are looking for a maximum for $\displaystyle I$, we take its derivative and set it equal to zero, and solve for $\displaystyle t$:

$\displaystyle \frac{dI}{dt} = \frac{d}{dt} \left[ te^{-\alpha t} \right] = e^{-\alpha t} - \alpha t e^{-\alpha t} = (1-\alpha t)e^{-\alpha t} = 0$

Since $\displaystyle e^{-\alpha t}$ can never equal zero, we know that:

$\displaystyle 1-\alpha t = 0 \implies t = \frac{1}{\alpha}$

So the value of $\displaystyle t$ where the maximum occurs is $\displaystyle t=\frac{1}{\alpha}$. Therefore:

$\displaystyle I = t e^{-\alpha t} = \frac{1}{\alpha} e^{-1} = \frac{1}{\alpha e} \approx 10$