A rectangle has its base on the x-axis and its 2 upper corners on the parabola y=12-x^2. What is the largest possible area of the rectangle?

Since this is an upside down parabola, the rectangle must be inside the parabola. The parabola and rectangle inside it are symmetrical about the y axis.

Let x be the coordinate of a corner that touches the parabola at y = 12 - x^2. The area of that parabola is
A = 2x*(12-x^2)= 24 x - 2x^3
That is a maximum when dA/dx = 0
24 -6x^2 = 0
x = +/- 2
y = 12 - 4 = 8
Maximum area = 2xy = 32

To find the largest possible area of the rectangle, we need to determine the dimensions that maximize the area. Let's break down the problem into steps:

1. We are given that the base of the rectangle lies on the x-axis. This means that the length of the rectangle is equal to the distance between the x-coordinates of the upper corners of the rectangle.

2. The two upper corners of the rectangle lie on the parabola y = 12 - x^2. We can find the x-coordinates of these points by setting y = 12 - x^2 equal to zero and solving for x.

12 - x^2 = 0
x^2 = 12
x = ±√12
x = ±2√3

So the x-coordinates of the upper corners are x = -2√3 and x = 2√3.

3. The length of the rectangle is the difference between the x-coordinates of the upper corners:
Length = 2√3 - (-2√3) = 4√3

4. The width of the rectangle is the y-coordinate of the upper corners. We can find this by substituting the x-coordinate into the equation y = 12 - x^2:

For x = -2√3:
y = 12 - (-2√3)^2 = 12 - 12 = 0

For x = 2√3:
y = 12 - (2√3)^2 = 12 - 12 = 0

The width of the rectangle is 0.

5. The area of the rectangle is given by the formula: Area = Length * Width.
Area = 4√3 * 0 = 0

Therefore, the largest possible area of the rectangle is 0.

To find the largest possible area of the rectangle, we need to find the dimensions (length and width) that maximize the area.

Let's start by drawing a diagram to visualize the problem:

```
^
| 12
| .
. . . . . . . . . . . . . . . . . . . . . .
|
|
|_______ x-axis
```

We have a rectangle with its base on the x-axis, and its upper corners on the parabola y = 12 - x^2. Let's assume the length of the rectangle is 2x, and the width is 2y.

The area of the rectangle is given by: A = length * width = (2x) * (2y) = 4xy.

To find the largest possible area, we need to maximize this function: A = 4xy.

Now, we need to express one variable in terms of the other variable. Since the upper corners of the rectangle lie on the parabola y = 12 - x^2, we can substitute this y-value into the area equation:

A = 4x(12 - x^2).

Next, we need to take the derivative of A with respect to x and set it equal to zero to find the critical points (points where the derivative is zero or undefined):

dA/dx = 4(12 - x^2) - 8x(2x) = 48 - 4x^2 - 16x^2 = 48 - 20x^2 - 16x = 0.

Simplifying the equation, we get:

-20x^2 - 16x + 48 = 0.

To solve this quadratic equation, we can use factoring, completing the square, or the quadratic formula. Let's use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a).

In our case, a = -20, b = -16, and c = 48. Substituting these values into the formula:

x = (-(-16) ± √((-16)^2 - 4(-20)(48))) / (2(-20)).

Simplifying further, we get:

x = (16 ± √(256 + 3840)) / (-40).

Calculating the square root and simplifying:

x = (16 ± √(4096)) / (-40)
x = (16 ± 64) / (-40).

So we have two possible x-values:

1) x = (16 + 64) / (-40) = 80 / (-40) = -2.
2) x = (16 - 64) / (-40) = -48 / (-40) = 1.2.

Since we are dealing with a rectangle on the x-axis, we can discard the negative value of x (-2) as it would not make sense in this context.

Now, to find the corresponding y-value, we substitute the x-value (1.2) back into the equation y = 12 - x^2:

y = 12 - (1.2)^2 = 12 - 1.44 = 10.56.

Therefore, the dimensions of the rectangle that maximize the area are:
Length: 2x = 2 * 1.2 = 2.4
Width: 2y = 2 * 10.56 = 21.12

The largest possible area of the rectangle is the product of the length and width:

Area = length * width = 2.4 * 21.12 = 50.688.