JEE PYQ: Parabola Question 31
Question 31 - 2019 (12 Jan Shift 2)
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:
(1) $36$
(2) $20\sqrt{2}$
(3) $32$
(4) $18\sqrt{3}$
Show Answer
Answer: (3)
Solution
Rectangle vertices $(\pm t, 0)$ and $(\pm t, 12 - t^2)$. Area $= 2t(12 - t^2) = 24t - 2t^3$. $\frac{dA}{dt} = 24 - 6t^2 = 0 \Rightarrow t = 2$. Maximum area $= 48 - 16 = 32$.
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