A rectangle with its base on the x-axis is to be inscribed under the graph of y=2-x^2. Find the height of the rectangle if the area is the largest possible area.
let the point of contact of the rectangle with the parabola in the first quadrant be P(x,y)
So the base of the rectangle is 2x and its height is y
Area = 2xy
= 2x(2-x^2)
= 4x = 2x^3
d(Area)/dx = 4 - 6x^2 = 0 for a max/min of area
6x^2 = 4
x = √(2/3)
then the height for a max area
= 2 - 2/3
= 4/3
To find the maximum area of the rectangle, we need to determine the dimensions of the rectangle that will maximize its area.
Let's start by visualizing the problem. We have a rectangle that is inscribed under the graph of y = 2 - x^2, with its base on the x-axis. The height of the rectangle is parallel to the y-axis, and the width is parallel to the x-axis.
To find the maximum area, we can express the area of the rectangle as a function of one variable, and then find the maximum of that function using calculus.
Let's denote the width of the rectangle as "w" and the height as "h". Since the base of the rectangle is on the x-axis, the width of the rectangle will be the difference between the x-coordinate of the two corners.
The left corner of the rectangle will have an x-coordinate of -w/2, and the right corner will have an x-coordinate of w/2. Since the rectangle is inscribed under y = 2 - x^2, the heights of the two corners will be given by the equation y = 2 - x^2.
So, the height of the rectangle, "h," will be the difference between the y-coordinates of the two corners:
h = (2 - x^2) - (2 - (-x^2)) = 2 - x^2 + x^2 = 2.
Therefore, the height of the rectangle is fixed at h = 2.
Now, we can express the area (A) of the rectangle as a function of the width (w):
A = w * h = w * 2 = 2w.
To maximize the area, we need to find the maximum value of the function A(w) = 2w.
Since A(w) is a linear function, it increases indefinitely as w increases. Therefore, there is no maximum area for the rectangle under the given conditions.
In summary, the height of the rectangle is fixed at 2, but there is no maximum area for the rectangle as it increases indefinitely with the width.