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JEE Main 2019 · 12 Jan · Shift I

JEE Main 2019 Mathematics question on Application of Derivatives

The maximum area (in sq. units) of a rectangle having its base on the XX-axis and its other two vertices on the parabola, y=12−x2y = 12 - x^2 such that the rectangle lies inside the parabola, is
  1. A.
    36
  2. B.
    20220\sqrt{2}
  3. C.
    32
  4. D.
    18318\sqrt{3}
Show answer and solution

Correct option: C

Solution

Let the rectangle have vertices at (±x, 0) on the x-axis and (±x, y) on the parabola where y = 12 - x².

Area A = 2x · y = 2x(12 - x²) = 24x - 2x³.

dA/dx = 24 - 6x² = 0 → x² = 4 → x = 2.

Maximum area = 2(2)(12 - 4) = 4 · 8 = 32.

[JEE Main 2019 · 12 Jan · Shift I]

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