Question

Find the area of the largest rectangle having two vertices on the x axis and two more vertices above the x axis and on the parabola with equation y=9-x²?

Answer 1

`A_max = 12sqrt3`

Explanation:

For reasons of symmetry, the rectangle will be symmetric with respect to the `y` axis and thus the two vertices lying on the `x` axis have coordinates `(-x,0)` and `(x,0)` and the ones lying on the parabola have coordinates `(-x, 9-x^2)` and `(x,9-x^2)`, and because they must lie above the `x` axis we can consider only `x in [0,3)`.

We have to maximize the area:

`A(x) = 2x(9-x^2)` for `x in [0,3)`

`A(x) = 18x -2x^3`

`(dA)/dx = 18-6x^2 = 6(3-x^2)`

To find the critical points we equate the derivative to zero:

`(dA)/dx = 0 => x=sqrt(3)`

and choose the solution in the interval `[0,3)`

`(d^2A)/dx^2 = -9x`

As for `x=sqrt3`, we have that `(d^2A)/dx^2 <0`, the critical point is a maximum and the the maximum area is:

`A(sqrt3) = 2sqrt3(9-3) = 12sqrt3`