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. unitsYou need to login to perform this action.
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