Area of largest triangle under $y=e^{-x}$ I came across this question and I'm unsure how to solve it. The question wasn't originally in English, and English isn't my first language so please excuse any terminology/grammar mistakes. 

A tangent is drawn through the point P on the curve $y=e^{-x}$. The tangent together with the positive $y$-axis and a horizontal line that goes through P cuts a triangle. Find the biggest possible value of the triangle's area.

Unfortunately I wasn't able to add a picture.
The placement of P is very important to this question, but I can't figure out where its most logical placement would be.
I know I can find the equation of the tangent by $y-f(x_0)=f'(x_0)(x-x_0)$ but I'm unsure how to go from there. 

Let $P(\alpha, e^{-\alpha})$ be the point then the tangent
$y-e^{-\alpha}=-e^{-\alpha}(x-\alpha)$
meets Y-axis at $A(0,e^{-\alpha}(\alpha +1))$. The horizontal line through $P$ meets Y-axis at $B(0,e^{-\alpha})$.
The area of triangle $PAB$ is 
$S(\alpha)=\frac{1}{2} (\alpha)^2 e^{-\alpha}$
Use derivatives to see that $\alpha= 0, 2$ are the stationary points. Also the second derivative $S''(2)<0$, so $S_{max}=2/e^2$
