Find Largest Rectangle on y=12-x^2

[tandoorichicken]

Find the area of the largest rectangle with lower base on the x-axis and upper vertices on the curve y = 12-x^2

 

[himanshu121]

Since one side is on x-axis then other side will be || x-axis(rectangle) let say y=a then u will x=+sqrt(12-b).

Therefore u have length say = 2sqrt(12-b) &
breadth = b.

Hence Area, $$A = b\sqrt{12-b}$$.

u will get b=8 and Length=8& breadth 8.

Alternate
Rectangle with max area is a square
u get $$b=2\sqrt{12-b}$$

 

[M98Ranger]

To find the largest rectangle on the curve y = 12-x^2, we first need to find the points of intersection between the curve and the x-axis. This can be done by setting y = 0 and solving for x.

0 = 12-x^2
x^2 = 12
x = ±√12
x = ±3.464

Therefore, the two points of intersection are (3.464, 0) and (-3.464, 0).

To find the area of the largest rectangle, we need to determine the height and width of the rectangle. The height of the rectangle will be the y-coordinate of the upper vertices, which is 12-x^2.

The width of the rectangle will be the distance between the two points of intersection, which is 2√12 or approximately 6.928.

Therefore, the area of the largest rectangle is:

A = height * width
A = (12-x^2) * 6.928
A = (12-12) * 6.928
A = 0 * 6.928
A = 0

Since the area of the rectangle is 0, this means that the largest rectangle on the curve y = 12-x^2 is a degenerate rectangle, meaning it is essentially a line segment with no width.

In other words, there is no largest rectangle on this curve that has a positive area. This is because as the width of the rectangle increases, the height decreases, resulting in a smaller area. And as the width decreases, the height increases, but the area still remains 0.

Therefore, the answer to this question would be that there is no largest rectangle on the curve y = 12-x^2.