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 - x 2 such that the rectangle lies inside the parabola, is :

Step-by-step solution

Since, given parabola is symmetric about the y - axis, hence rectangle will also be symmetric about y - axis. Let one vertex of the rectangle on the x - axis be α , 0 , then Area of rectangle A = 2 α . 12 - α 2 Differentiating both sides with respect to α , we get ⇒ d A d α = 24 - 6 α 2 = 0 ⇒ α = 2 ,   - 2 For area to be maximum, put α = 2 in the equation for area of the rectangle, we get A m a x = 2 × 2 × 12 - 2 2 = 32