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 :

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Aicharya · Accepted Answer

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

$\Rightarrow \frac{d A}{d α} = 24 - 6 α^{2} = 0 \Rightarrow α = 2, - 2$

For area to be maximum, put $α = 2$ in the equation for area of the rectangle, we get

$A_{m a x} = 2 \times 2 \times (12 - 2^{2}) = 32$

The maximum area (in sq. units) of a rectangle having its base on the x - axis and its other two vertices on…

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