Question

A rectangle is constructed with it's base on the x-axis and the two of its vertices on the parabola `y=49 - x^2`. What are the dimensions of the rectangle with the maximum area?

Answer 1

In other words, we're constructing a rectangle under a dome-shaped form.

Explanation:

The parabola is a 'mountain'-type (because the coefficient of `x^2` is negative. Also, it is symmetrical in respect to the `y`-axis, because there is no `x`-term.
We can now simplify the problem to finding a rectangle with vertices at `(0,0) , (x,0), (0,y) and (x,y)` and then double the `x`-values.
The area will then be `A=x*y`
If we substitute the equation of the parabola for `y`:
`A=x*(49-x^2)=49x-x^3`

To find the extremes (max of min) we need the derivative and set it to zero:
`A'=49-3x^2=0->x^2=49/3->x=sqrt(49/3)~~4.04...`
(remember we will have to double that)

Use this in the original function:

`y=49-x^2=49-49/3=98/3~~32.67`

Answer :
Dimensions will be `8.08" x "32.67`
graph{49-x^2 [-65.4, 66.33, -13.54, 52.3]}