What is the area of the largest rectangle with lower base on the x-axis and upper vertices on the curve y = 12 - x^2?
To find the area of the largest rectangle with a lower base on the x-axis and upper vertices on the curve y = 12 - x^2, we can employ calculus and optimization techniques.
The first step is to establish the equation of the rectangle. Since the base of the rectangle lies on the x-axis, the length of the rectangle will be equal to the difference between the x-coordinates of the upper vertices. Let's denote this length as 2x, as it extends from x to -x on the x-axis.
Now, to determine the height of the rectangle, we need to find the corresponding y-values on the curve y = 12 - x^2. Since the upper vertices of the rectangle lie on this curve, the height is given by the function itself, which is y = 12 - x^2.
The area of the rectangle is then given by the product of the length (2x) and the height (12 - x^2):
A = 2x * (12 - x^2)
To find the maximum area, we need to differentiate this expression with respect to x and set it equal to zero. Then we can solve for the value of x that maximizes the area.
dA/dx = 0
Differentiating the equation and solving for x:
dA/dx = 2(12 - x^2) - 2x(-2x) = 0
24 - 2x^2 + 4x^2 = 0
2x^2 = 24
x^2 = 12
x = sqrt(12)
Having found the value of x, we can calculate the maximum area by substituting this value back into the area formula:
A = 2(sqrt(12)) * (12 - sqrt(12)^2)
A = 2(sqrt(12)) * (12 - 12)
A = 2(sqrt(12)) * 0
A = 0
Therefore, the area of the largest rectangle with a lower base on the x-axis and upper vertices on the curve y = 12 - x^2 is zero. This implies that there is no rectangle that meets the given criteria.