Find the maximum area of a rectangle that lies above the x axis and below y=9-x^2

any candidate rectangle must have its vertices on the x-axis and on the curve.

If (x,y) is a vertex, so is (-x,y)

A = 2x(9-x^2) = 18x - 2x^3
dA/dx = 18 - 6x^2
A' = 0 when x = √3

A(√3) = 2√3(9-3) = 12√3

To find the maximum area of a rectangle that lies above the x-axis and below the curve y = 9 - x^2, we can use calculus.

The area of a rectangle is given by the formula A = length * width. In this case, the length of the rectangle will be along the x-axis, and the width will be along the y-axis.

Let's assume that the length of the rectangle is 2x (twice the x-coordinate) and the width is y (the y-coordinate). We want to maximize A = 2x * y.

To proceed, we need to express y in terms of x. Since the rectangle lies below the curve y = 9 - x^2, we have:

y = 9 - x^2

Now let's substitute this expression for y in the area formula:

A = 2x * (9 - x^2)

Expanding the expression, we get:

A = 18x - 2x^3

To find the maximum area, we need to find the critical points where the derivative of A with respect to x is zero or does not exist.

First, let's find the derivative of A with respect to x:

dA/dx = 18 - 6x^2

Setting dA/dx equal to zero and solving for x:

18 - 6x^2 = 0
6x^2 = 18
x^2 = 3
x = ±sqrt(3)

Now, we need to check if these critical points correspond to a maximum or minimum. To determine this, we can find the second derivative of A with respect to x:

d^2A/dx^2 = -12x

For x = sqrt(3):

d^2A/dx^2 = -12 * sqrt(3) < 0

For x = -sqrt(3):

d^2A/dx^2 = -12 * (-sqrt(3)) = 12 * sqrt(3) > 0

This tells us that x = -sqrt(3) corresponds to a local maximum, and x = sqrt(3) corresponds to a local minimum. We are interested in the maximum area, so we will consider x = -sqrt(3).

Substituting x = -sqrt(3) into the area formula, we get:

A = 18(-sqrt(3)) - 2(-sqrt(3))^3
A = -18sqrt(3) - 2(-3sqrt(3))
A = -18sqrt(3) + 6sqrt(3)
A = -12sqrt(3)

Since we are looking for the maximum area, we take the absolute value of -12sqrt(3):

Max Area = | -12sqrt(3) |

Therefore, the maximum area of the rectangle is 12sqrt(3) square units.