The maximum area (in sq. units) of a rectangle having its base on the x-axis and its other two vertices on the parabola, y = 12 - x², such that the rectangle lies inside the parabola, is :

Let the rectangle have vertices at (-a, 0), (a, 0), (-a, 12 - a²), and (a, 12 - a²), where a > 0.
The length of the rectangle is 2a and the width is 12 - a².
The area of the rectangle is given by A(a) = 2a(12 - a²) = 24a - 2a³.
To find the maximum area, we take the derivative of A(a) with respect to a and set it to zero:
A'(a) = 24 - 6a² = 0
6a² = 24
a² = 4
a = 2 (since a > 0)
Now we find the second derivative to check if this is a maximum:
A''(a) = -12a
A''(2) = -24 < 0, so this is a maximum.
The maximum area is A(2) = 2(2)(12 - 2²) = 4(12 - 4) = 4(8) = 32 square units.