A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola y+1-x^2. What are the dimensions of such a rectangle with the greatest possible area?

width? height?

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

Given that the base of the rectangle is on the x-axis, the length of the rectangle is equal to the x-coordinate of the upper corners of the rectangle. Let's denote the x-coordinate as x.

The y-coordinate of the upper corners of the rectangle is given by the parabola equation y + 1 = x^2. By rearranging the equation, we have y = x^2 - 1.

Since the rectangle is inscribed within the parabola, the height of the rectangle is calculated by finding the difference between the y-coordinates of the upper corners and the x-axis. Therefore, the height (h) of the rectangle is equal to y.

Now, we can express the area (A) of the rectangle in terms of x and y: A = x × y.

To maximize the area, we need to find the maximum value of A. Let's substitute y with x^2 - 1 in the equation above:

A = x × (x^2 - 1)

To find the maximum value of A, we can take the derivative of A with respect to x, set it equal to zero, and solve for x:

dA/dx = (x^2 - 1) + x(2x) = 0
x^2 - 1 + 2x^2 = 0
3x^2 - 1 = 0
3x^2 = 1
x^2 = 1/3
x = ±√(1/3)

Since the base of the rectangle is on the x-axis, x cannot be negative. Therefore, x = √(1/3).

Now, we can find the height (h) by substituting x into the equation y = x^2 - 1:

y = (√(1/3))^2 - 1 = 1/3 - 1 = -2/3

Since the height cannot be negative, we take the absolute value:

h = | -2/3 | = 2/3

Therefore, the width of the rectangle is 2 × (√(1/3)) = 2√(1/3) and the height is 2/3.