Question

Let P(x,y) be a point on the graph of y=−x^2+8 with 0<x<√8. Let PQRS be a rectangle with one side on the x-axis and two vertices on the graph. Find the rectangle with the greatest possible area. Enter the exact value of the area of this rectangle.

Greatest area: ___unit^2

Answer 1

In the standard version of this question , let P(x,y) be the point in quadrant 1
then for the rectangle,
the length of the base will be 2x, and the height is y

Area = 2xy = 2x(-x^2 + 8) = 16x - 2x^3
d(Area)/dx = 16 - 6x^2 = 0 for a max/min of Area
6x^2 = 16
x^2 = 16/6
x = 4/√6 = 2√6 / 3
x = 2√6/3 , then y = -(16/6)+8 = 16/3

area = 2(2√6/3)(16/3) = 64√6/9

check my arithmetic

Answer 2

It says that the answer is incorrect.

Answer 3

Sorry, that answer is correct.
I ran a computer simulation program and got the same result
How did you enter the answer, did you use the √ sign?

I also rationalized the denominators, try entering the results
with square roots in the denominator.