find the the area of the largest rectangle that can be under the parabola y=9-x^2 above the x axis?
Let the base of the rectangle have length 2x.
Then the area of the rectangle is
a = 2xy = 2x(9-x^2)
da/dx = 6(3-x^2)
so da/dx = 0 when x = ±√3
we want x>0, so a has a max of 12√3
To find the area of the largest rectangle that can be under the parabola y = 9 - x^2 above the x-axis, we need to find the dimensions of the rectangle first.
Let's assume that the rectangle has sides parallel to the x and y axes and that its base lies along the x-axis. We will find the height of the rectangle and then calculate its area.
To begin, we need to find the x-coordinate of the vertex of the parabola y = 9 - x^2, as this will divide the parabola symmetrically.
The x-coordinate of the vertex can be found using the formula x = -b / (2a), where a and b are the coefficients of x^2 and x, respectively, in the equation y = ax^2 + bx + c. In this case, a = -1 and b = 0 (since there is no x-term). So, we have:
x = -0 / (2(-1))
x = 0 / 0 (which is undefined)
Since the x-coordinate is not defined, it means that the vertex of the parabola lies at the y-intercept. In this case, the y-intercept of the parabola y = 9 - x^2 is (0, 9).
Now, for the rectangle to be the largest possible area under the parabola, it should have its top side touching the curve of the parabola. Therefore, the height of the rectangle will be the y-coordinate of the top side, which is 9.
Since the base of the rectangle is along the x-axis, its length will be equal to the distance between the x-intercepts of the parabola.
To find the x-intercepts, we set y = 0 in the equation y = 9 - x^2 and solve for x:
0 = 9 - x^2
x^2 = 9
x = ±√9
x = ±3
Therefore, the length of the base of the rectangle is 6 units (2 * 3).
Now that we have the height and base of the rectangle, we can calculate its area using the formula: Area = height * base.
Area = 9 * 6
Area = 54 square units.
Therefore, the area of the largest rectangle that can be under the parabola y = 9 - x^2 above the x-axis is 54 square units.
To find the area of the largest rectangle under the parabola y = 9 - x^2 above the x-axis, we need to maximize the area of the rectangle. This can be done by finding the maximum length and maximum width, then multiplying them together.
Let's start by visualizing the problem. The parabola y = 9 - x^2 opens downwards and intersects the x-axis at x = -3 and x = 3. The area of the rectangle will be bounded by the points of intersection.
To find the maximum length, we need to find the distance between the points of intersection. In this case, the distance is 2 units (from -3 to 3). Therefore, the maximum length of the rectangle is 2.
To find the maximum width, we need to find the maximum height of the rectangle. The height will be determined by the y-values of the parabola. The vertex of the parabola is at (0, 9), which is the maximum y-value. Therefore, the maximum height of the rectangle is 9.
Now we can calculate the area by multiplying the maximum length (2) by the maximum width (9):
Area = Length × Width = 2 × 9 = 18 square units.
Hence, the area of the largest rectangle that can be formed under the parabola y = 9 - x^2 above the x-axis is 18 square units.