Question

Find the area A of the largest rectangle that can be inscribed under the curve of the equation below in the first and second quadrants.

y = e−x2

Answer 1

Let
`(x,0)` with
`x>0` be the lower right vertex of the rectangle, so its base has total length
`2x`. The height of the rectangle will be the function value this value of
`x`, or
`e^(-x^2)`.

Then the area of the rectangle is

`A(x) = 2 x e^(-x^2)`

Differentiate
`A` with respect to
`x` and find the critical points.

`A'(x) = 2 e^(-x^2) - 4x^2 e^(-x^2) = 2e^(-x^2) (1 - 2x^2)`

`A'(x) = 0 \implies 1 - 2x^2 = 0 \implies x = \frac1{\sqrt2}`

Evaluate the second derivative at this critical point to ensure
`A` is maximized there.

`A''(x) = -4xe^(-x^2) (1 - 2x^2) + 2e^(-x^2) (-4x) = 4e^(-x^2) (2x^3 - 3x)`

Since
`A''\left(\frac1{\sqrt2}\right) < 0`, this is indeed a local maximum. Then the largest area is

`A\left(\frac1{\sqrt2}\right) = \frac2{\sqrt2} e^(-(1/\sqrt2)^2) = \boxed{√(\frac2e)}\approx 0.8578`