Maximizing area under $y=e^{−{∣x∣}}$ The coordinates of the point $M(x,y)$ on $y=e^{−{∣x∣}}$ so that the area formed by the coordinates axes and the tangent at $M$ is greatest is what?
I tried to plot the graph but after that I'm not being able to proceed.Please help]1
 A: the problem is symmetric so will look at the tangent at $P=(1, 1/e^a), a > 0$ and the area of the region bounded by the tangent, positive $x$-axis and the positive $y$-axis.
the tangent at  $P$ has slope $-\frac1{e^a}$ and is given by $$y - \frac1{e^a}=-\frac1{e^a}\left( x - a \right) $$ the $x$-intercept is $(1+a, 0)$ and the $y$-intercept is $\frac{1+a}{e^a}.$ so the required area is $$A = \frac{(1+a)^2}{2e^a}, 0 \le a < \infty \tag 1$$ 
you can verify that $A$ has a global max of $\frac 2 e$ at $a = 1.$   
A: The graph is symmetric around the $y$-axis, so if $M(x,y)$ is an optimal solution, so will $M(-x,y)$ be as well. With that in mind, let's look for the solution in the I quadrant, so assume $x \ge 0$.


*

*There your curve looks like $y = e^{-x}$. Find the equation of the tangent.

*At arbitrary $x$, what is the slope of the tangent line? What are it's intercepts with each axis?

*What is the area of the enclosed triangle?


Maximize this area as a function of $x$...
UPDATE
So the curve we are looking at is given by $f(x) = e^{-x}$ for $x \ge 0$. Note that $f'(x) = -e^{-x}$.
At an arbitrary point $(u,v)$ with $x = u$, we have $v = f(u) = e^{-u}$ and the slope of the tangent line is $f'(u) = -e^{-u}$. Hence, the equation of the tangent line is given by
$$y - v = -e^{-u} (x-u).$$
Since $v = e^{-u}$ we have
$$y = v -e^{-u} (x-u) = e^{-u} -e^{-u} (x-u) = e^{-u}[1-x+u],$$
which crosses the $y$-axis at $e^{-u}[1+u]$ and the $x$-axis at the solution of $$0 = e^{-u}[1-x+u],$$
which must be $x = 1+u$, since $e^{-u} > 0$.
Can you please finish the problem andlet me know what you get?
