Answer
The rectangle has the largest possible area when the vertices are at the points of inflection of the curve.
Work Step by Step
Let $(x,y)$ be the point in the first quadrant where the rectangle has a vertex that intersects the curve $y = e^{-x^2}$
We can write an expression for the area of the rectangle:
$A = (2x)(y)$
$A = (2x)(e^{-x^2})$
We can find $\frac{dA}{dx}$:
$\frac{dA}{dx} = 2e^{-x^2}-4x^2~e^{-x^2} = 0$
$(2-4x^2)~e^{-x^2} = 0$
$2-4x^2 = 0$
$4x^2 = 2$
$x^2 = \frac{1}{2}$
$x = \pm \sqrt{\frac{1}{2}}$
The rectangle has the largest possible area when the x-coordinates of the vertices are $~~x = -\sqrt{\frac{1}{2}}~~$ and $~~x = \sqrt{\frac{1}{2}}$
We can find the x-coordinates of the points of inflection:
$y = e^{-x^2}$
$y' = -2x~e^{-x^2}$
$y' = -2~e^{-x^2}+4x^2~e^{-x^2} = 0$
$(-2+4x^2)~e^{-x^2} = 0$
$2-4x^2 = 0$
$4x^2 = 2$
$x^2 = \frac{1}{2}$
$x = \pm \sqrt{\frac{1}{2}}$
The x-coordinates of the points of inflection are $~~x = -\sqrt{\frac{1}{2}}~~$ and $~~x = \sqrt{\frac{1}{2}}$
The rectangle has the largest possible area when the vertices are at the points of inflection of the curve.
