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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 [JEE Main Online Paper Held On 12-Jan-2019 Morning]

A) 36                                

B) \[18\sqrt{3}\]

C) \[20\sqrt{2}\]                

D)   32

Correct Answer: D

Solution :

Let (a, 0) be any point on the x-axis, which is the vertex of the rectangle So, the co-ordinates of the vertex of the rectangle lying on the parabola \[y=12-{{x}^{2}}\]is\[(a,12-{{a}^{2}}).\] \[\therefore \]Area of rectangle,\[f(a)=2a(12-{{a}^{2}})\] \[\therefore \] \[f'(a)=2(12-3{{a}^{2}})\] For maximum area\[f'(a)=0\]\[\Rightarrow \]\[2(12-3{{a}^{2}})=0\] \[\Rightarrow \]\[a=\pm 2\]   \[\therefore \]Maximum area at a=2 is f(2)=32 sq. units