Find the largest possible area for a rectangle with base on the x-axis and upper vertices on the curve

y=8-x^2 find the largest possible area for a rectangle with base on the x-axis and upper vertices on the curve

y=8-x^2

So here is how we find the answer to this given problem.
The formula for area is: A=(l)(w)
So, A=(2x)(8-x²)
A=16x-2x³
dA/dx = 16-6x²
then you will use calculus to get the absolute local maximum of x by letting dA/dx = 0
then wha you will get is x = 2sqrt(6)/3
so the area would be= 64 * sqrt(6) / 9
or 64√6/9.
Based on the choices given, the answer would be option D. 
Hope this answer helps.

4 units

Step-by-step explanation:

Since the two triangles are similar, their sides are proportional. To find the length of the side, you need to use the distance formula:

The coordinates you will use are (0, 0) and (-4, 0) because they correspond to points X and Y in Triangle PQR:

Then, simplify everything under the radical:

Finally, take the square root of 16:

= 4