A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola y=4−x^2. What are the dimensions of such a rectangle with the greatest possible area?

Width =
Height =

If the base has width 2x and height y, then the area is

a = 2xy = 2x(4-x^2)

so find x when da/dx =0

To find the dimensions of the rectangle with the greatest possible area, we need to maximize the area function of the rectangle. The area of a rectangle is given by length × width.

First, let's set up the problem. We have a rectangle inscribed with its base on the x-axis, so the length of the rectangle will be equal to the height of the parabola function.

The equation of the parabola is y = 4 - x^2. To find the height of the rectangle, we need to determine the y-coordinate of the upper corners of the rectangle. Since the base of the rectangle is on the x-axis, the y-coordinate of these upper corners will be equal to the value of the parabola function at the x-coordinate of the corners. Let's call these x-coordinates A and B.

So, we have two points on the parabola: A (A, 4 - A^2) and B (B, 4 - B^2). The length of the rectangle (which is equal to the height of the parabola) will be the difference between the y-coordinates of the two points: (4 - A^2) - (4 - B^2) = B^2 - A^2.

The width of the rectangle is simply the difference between the x-coordinates of the corners: B - A.

To find the dimensions of the rectangle with the greatest possible area, we need to maximize the area function: Area = length × width = (B^2 - A^2)(B - A).

Since the area function is a product of two terms, we can use calculus to find its maximum value. Taking the derivative of the area function with respect to A and setting it equal to zero, we can find the critical points which correspond to maximum area.

d(Area)/dA = -2A(B - A) + (B^2 - A^2)(-1) = 0

Simplifying the equation, we get:

2A(B - A) = B^2 - A^2

2AB - 2A^2 = B^2 - A^2

2AB = B^2

Simplifying further, we get:

2A = B

Now we have a relation between A and B. Substituting this relation into one of the original equations, we can solve for A and B.

Since 2A = B, we can substitute A with (B/2) in the equation y = 4 - x^2:

y = 4 - (B/2)^2

Simplifying the equation, we get:

y = 4 - (B^2/4)

Setting this equation equal to zero and solving for B, we get:

4 - (B^2/4) = 0

B^2 = 16

B = ±4

Since B represents the x-coordinate of the upper corners of the rectangle, we only need to consider positive values of B.

So, B = 4.

Using the relation 2A = B, we can solve for A:

2A = 4

A = 2

Therefore, the dimensions of the rectangle with the greatest possible area are:

Width = B - A = 4 - 2 = 2
Height = (4 - A^2) - (4 - B^2) = (4 - 2^2) - (4 - 4^2) = 0 - 0 = 0

Hence, the rectangle is a degenerate rectangle with zero height.