Question

A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola `y=6-x^2`. What are the dimensions of such a rectangle with the greatest possible area? thanks for any help!?

Answer 1

Height: `4` units
Width `2sqrt(2)`

Explanation:

Start by sketching `y=6 -x^2`. Then draw a rectangle beneath it. You will notice that the width is `2x` and the height is `6-x^2`. Area is given by length times width, so the area function will be `A=2x(6-x^2) =12x - 2x^3`.

Now you differentiate to find the maximum.

`A’ =12 -6x^2`

Find critical numbers by setting `A’` to `0`.

`x =+-sqrt(2)`

The derivative is negative at `x=2` and positive at `x=1`, which justifies that the rectangle with width of `sqrt(2)` has maximal area.

The height will be `y(sqrt(2)) = 6-(sqrt(2))^2) = 4`

Hopefully this helps!