A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola f(x)=8-x^2. What are the dimensions of such a rectangle with the greatest possible area?
Let the base of the rectangle go from -x to x. The area of the rectangle is thus
a = 2xy = 2x(8-x^2)
da/dx = 16 - 6x^2
da/dx = 0 when x = 4/√6
So, the rectangle is 8/√6 by 16/3
To find the dimensions of the rectangle with the greatest possible area, we need to maximize the area of the rectangle within the given constraints.
Let's start by visualizing the problem. The base of the rectangle is on the x-axis, so it will have a width w. The height of the rectangle will be equal to the y-coordinate of the points on the parabola above the x-axis.
Since the upper corners of the rectangle lie on the parabola f(x) = 8 - x^2, we can find the y-coordinate of these points by substituting x into the equation.
Given that the base of the rectangle is the x-axis, the upper corners of the rectangle will have coordinates at (x, 8 - x^2). Therefore, the height of the rectangle will be 8 - x^2.
The area of a rectangle is given by the formula A = width * height, so the area of the rectangle is A = w * (8 - x^2).
To maximize this area, we need to find the values of w and x that will give us the maximum value for A.
Since the base of the rectangle is on the x-axis, the width w is equal to the distance between the x-coordinates of the two upper corners. Let's call these x-coordinates x1 and x2.
Therefore, the width of the rectangle is w = x2 - x1.
To find the maximum area, we need to find the maximum value for A = w * (8 - x^2) by determining the optimal values for w, x1, and x2.
Taking the derivative of A with respect to x and setting it equal to zero will help us find the critical points where the area is maximized.
dA/dx = 0
0 = w * (d(8 - x^2)/dx)
Since we are looking for the maximum area, we can assume the width w is positive, so we can ignore it for now.
0 = d(8 - x^2)/dx
To evaluate this derivative, we apply the power rule:
0 = -2x
Solving for x gives us x = 0.
Now, let's determine the endpoints of the x-interval where we need to check for the maximum area.
The parabola f(x) = 8 - x^2 intersects the x-axis at (x, 0), so we need to find the x-values where f(x) = 0.
0 = 8 - x^2
x^2 = 8
x = ±√8
Since the rectangle is inscribed, x1 and x2 must lie between -√8 and √8, so the x-interval is -√8 ≤ x ≤ √8.
Now we have the interval and the critical point x = 0. We can evaluate the function A(x) on this interval and at the critical point to find the maximum area.
A(x) = w * (8 - x^2)
First, let's check the endpoints of the interval:
A(-√8) = (√8 - (-√8)) * (8 - (-√8)^2) = 4√8 * (8 - 8) = 0
A(√8) = (√8 - √8) * (8 - (√8)^2) = 0
Now, let's check the critical point:
A(0) = (√8 - (-√8)) * (8 - 0^2) = 2√8 * 8 = 16√8
Therefore, the maximum area of the rectangle is 16√8, and it is achieved when x = 0.
To find the dimensions of the rectangle with the greatest possible area, we need to find the corresponding width w and height h.
Since the base of the rectangle is on the x-axis, the width w is equal to the distance between the x-coordinates of the two upper corners. In this case, since x1 = x2 = 0, the width w is also equal to zero.
The height h of the rectangle is given by the y-coordinate of the upper corners, which is f(x) = 8 - x^2. Substituting x = 0, we find that the height is h = f(0) = 8 - 0^2 = 8.
Therefore, the dimensions of the rectangle with the greatest possible area are width = 0 and height = 8.