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Question

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 :-


A
202
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B
183
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C
32
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D
36
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Solution

The correct option is C $$32$$
$$f ( a ) = 2 a ( 12 - a^2 )$$
$${ f }^{ { \prime  } }({ a })=2\left( 12-3{ a }^{ { 2 } } \right) $$
maximum at $$a = 2$$  
maximum area  $$= f ( 2 ) = 32$$
1142873_1331114_ans_0f53d84f87864535834a4ace288d4190.png

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