The base of a rectangle is on the x-axis and its two upper vertices are on the parabola y=25-x^2. of all such rectangles, what are the dimensions of the one with greatest area?
Clearly the maximum area is when the rectangle is centered on the parabola's axis of symmetry. In this case, x=0. So, let the base of the rectangle be of length 2x. Its area is thus
A = 2xy = 2x(25-x^2) = 50x-2x^3
dA/dx = 50-6x^2
dA/dx=0 when x=5/√3
So, the maximum area is (10/√3)(25-(5/√3)^2) = 500/(3√3)
To find the dimensions of the rectangle with the greatest area, we need to maximize the area of the rectangle. The area of a rectangle is given by the formula: Area = length * width.
Let's start by representing the rectangle on a coordinate plane. We know that the base of the rectangle is on the x-axis, so let's assume its length to be 2x. Therefore, the upper vertices of the rectangle will have coordinates (x, y) and (-x, y) on the parabola y = 25 - x^2.
Since these points lie on the parabola, we can substitute their x- and y-coordinates into the equation and solve for y.
For the first point (x, y):
y = 25 - x^2
For the second point (-x, y):
y = 25 - (-x)^2
= 25 - x^2
Since both points have the same y-coordinate, we can equate the two equations:
25 - x^2 = 25 - x^2
Simplifying, we get:
0 = 0
This implies that the y-coordinate of the upper vertices is the same regardless of the value of x.
Now, let's find the dimensions of the rectangle in terms of x. The length of the rectangle is given by 2x, as we assumed earlier, and the width is given by the distance between the y-coordinate of the upper vertices and the x-axis.
Since the y-coordinate of the upper vertices is always equal to 25, the width is given by 25 - 0 = 25.
Therefore, the dimensions of the rectangle with the greatest area are length = 2x and width = 25.