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
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
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.