I'm not sure if this was meant to be an unanswered question but here's a solution
Pick an arbitrary point (a,0) to be the first vertex of the rectangle. By default (a, e^(-a^2)) is also on the rectangle. But because it HAS TO BE A RECTANGLE and we have an even function
(-a,0) and (-a, e^(-a^2)) must lie on the rectangle. This is most easily shown with a diagram.
So the breadth and length of the triangle are:
Hence we can combine these to give an expression for area:
Set dA/dx=0 to maximise:
Reject x=0 though, because if x=0 we have no rectangle (we can't have a width of 0).
So just show that the remaining value gives a maxima using a table of values or the second derivative. Note that because the curve is an even function, the negative term can be ignored.
I would've used latex but I'm thoroughly lazy to on a phone (Hello, Jamon here, I went through and added LaTex
)