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

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

It says that the answer is incorrect.

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.

To find the rectangle with the greatest possible area, we need to consider the properties of the graph and the constraints given.

First, let's find the coordinates of point P. Since point P lies on the graph of y = -x^2 + 8, we can substitute the x-coordinate into the equation to find the corresponding y-coordinate.

Let P(x, y) be a point on the graph:
y = -x^2 + 8

Substituting x:
y = -(√8)^2 + 8
y = -8 + 8
y = 0

So, point P is (√8, 0), which lies on the x-axis.

Next, let's find the coordinates of point Q. Since point Q is one of the vertices of the rectangle, it must also lie on the graph of y = -x^2 + 8. However, it must also be on the x-axis.

Since the x-coordinate of point Q should be on the x-axis, it must be 0. We can substitute this into the equation to find the corresponding y-coordinate.

Let Q(0, y) be a point on the graph:
y = -(0)^2 + 8
y = 8

So, point Q is (0, 8), which lies on the x-axis.

Now we have two vertices of the rectangle: P(√8, 0) and Q(0, 8). To find the area of the rectangle, we need to calculate the length of one side, which is the distance between P and Q.

Using the distance formula:
Length = √((x2 - x1)^2 + (y2 - y1)^2)

Length = √((0 - √8)^2 + (8 - 0)^2)
Length = √((-√8)^2 + 8^2)
Length = √(8 + 64)
Length = √72
Length = 6√2

Since the rectangle has one side on the x-axis, the side opposite the x-axis will have the same length. Therefore, the area of the rectangle is the product of the length and the width.

Area = 6√2 * 6√2
Area = 36 * 2
Area = 72 square units

Therefore, the exact value of the area of the rectangle with the greatest possible area is 72 square units.