Consider all rectangles lying in the region $$\left\{(x, y) \in \mathbb{R} \times \mathbb{R}: 0 \leq x \leq \frac{\pi}{2} \text{ and } 0 \leq y \leq 2\sin(2x)\right\}$$ and having one side on the x-axis. The area of the rectangle which has the maximum perimeter among all such rectangles, is

JEE Advanced · 2020 · 27 Sep 2020 · Paper 1 · Application of Derivatives

Consider all rectangles lying in the region

{(x, y) \in R \times R : 0 \leq x \leq \frac{π}{2} and 0 \leq y \leq 2 sin (2 x)}

and having one side on the x-axis. The area of the rectangle which has the maximum perimeter among all such rectangles, is

Answer: (C)

By symmetry about

x = \frac{π}{4}

, take the rectangle with base from

\frac{π}{4} - t

\frac{π}{4} + t

and height

2 sin (2 (\frac{π}{4} - t)) = 2 cos 2 t

. Its perimeter is

P = 2 (2 t) + 2 (2 cos 2 t) = 4 t + 4 cos 2 t

, and

\frac{d P}{d t} = 4 - 8 sin 2 t = 0

gives

sin 2 t = \frac{1}{2}

, so

2 t = \frac{π}{6}

and

t = \frac{π}{12}

. The height is then

2 cos \frac{π}{6} = 3

and the base is

2 t = \frac{π}{6}

, so the area of this maximum-perimeter rectangle is

\frac{π}{6} \times 3 = \frac{π}{2 3}

JEE Advanced · 2020 · 27 Sep 2020 · Paper 1 · Single correct option