delete now plz
Let a be the point of the tangent line then
$\displaystyle (a,e^{-a})$ is a point on the graph
$\displaystyle y=e^{-x} \to y'=-e^{-x}$
So the slope at point a is
$\displaystyle m=-e^{-a}$
Now we can find the equation of the line through the point $\displaystyle (a,e^{-a})$
$\displaystyle y-e^{-a}=-e^{-a}(x-a) \iff y=-e^{-a}x+(a)e^{-a}+e^{-a} $
We can now find the x and y intercepts of the equation of the line
$\displaystyle (0,e^{-a}(a+1))$ and
$\displaystyle ((a+1),0)$
These are the lengths of the base and height of the triangle
So its area is
$\displaystyle A(a)=\frac{1}{2}(a+1)(e^{-a}(a+1))$
$\displaystyle A(a)=\frac{1}{2}(a+1)^2e^{-a}$
$\displaystyle \frac{dA}{da}=(a+1)e^{-a}-\frac{1}{2}(a+1)^2e^{-a}=e^{-a}(a+1)(1-\frac{1}{2}a-\frac{1}{2})$
setting each factor equal to zero and solving gives a=1 or a=-1
since we want the part in the first quadrant we don't use a=-1
$\displaystyle A(1)=\frac{1}{2}(1+1)(e^{-1}(1+1))=2e^{-1}=\frac{2}{e}$
Yeah!!!