JEE MAIN - Mathematics (2019 - 12th January Morning Slot - No. 7)
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 – x2 such that the rectangle lies inside the parabola, is :
36
20$$\sqrt 2 $$
18$$\sqrt 3 $$
32
Explanation
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f (a) = 2a(12 $$-$$ a)2
f '(a) = 2(12 $$-$$ 3a2)
Maximum at a = 2
maximum area = f(2) = 32
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