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. 
 A: The tangent will intersect the $y$-axis at the point 
$$
y-f(x_P)=[f'(x_P)(x-x_P)]_{x=0}=-f'(x_P)x_P.
$$
Thus the area of the triangle in question reads:
$$
A(x_P)=\frac{|f'(x_P)x_P^2|}2.
$$
Can you take it from here?
A: The derivative of the function is $f'(x)=-e^{-x}$
Let's say the point $P$ had the co-ordinates $(p, e^{-p}$
Then the tangent line at that point would be provided by:
$\dfrac{y-e^{-p}}{x-p}=-e^{-p} \implies y=-xe^{-p}+e^{-p}(p+1)$
We're talking about a positive y-axis so the lowest that line can go is the origin.
Turns out that that point satisfies the positive $y$ condition and maximizes both base and height of the triangle. So that would be when this happens:
$0=0 \cdot e^{-p}+e^{-p}(p+1)$
$e^{-p}(p+1)=0$
$p=-1$ or positive infinity. If you check that curve it tends to $0$ for bigger values of $x$. But that's not the point for now.
$P=(-1, e)$
There's the strategic point.
A: 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 $S_{max}=2/e^2$
