QJEEAdv20Maad63: JEE Advanced 2020 Mathematics - Class 12 JEEAdv20Maad63 Solution: Consider the rectangles lying the region and and having one…

Question

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Consider the rectangles lying the region

\left\{ {(x,y) \in R \times R:0\, \le \,x\, \le \,{\pi \over 2}} \right.

and

\left. {0\, \le \,y\, \le \,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

Accepted Answer

Given region is

\{ (x,y) \in R \times R:0\, \le \,x\, \le \,{\pi \over 2}

and

0\, \le \,y\, \le \,2\sin (2x)

On drawing the diagram,

Let the side PS on the X-axis, such that P(x, 0), and Q(x, 2sin(2x)), so length of the sides

PS = QR = 2\left( {{\pi \over 4} - x} \right)

and PQ = RS = 2sin 2x.

\therefore

Perimeter of the rectangle

y = 4\left[ {{\pi \over 4} - x + \sin 2x} \right]

For maximum,

{{dy} \over {dx}} = 0

\Rightarrow - 1 + 2\cos 2x = 0\, \Rightarrow \,\cos 2x = {1 \over 2}

\Rightarrow 2x = {\pi \over 3}\, \Rightarrow \,x = {\pi \over 6}\left\{ {x \in \left[ {0,\,{\pi \over 2}} \right]} \right\}

and

{\left. {{{{d^2}y} \over {d{x^2}}}} \right|_{x = {\pi \over 6}}} = {\left. { - 4\sin 2x} \right|_{x = {\pi \over 6}}}\, &lt; \,0

\therefore

At

x = {\pi \over 6}

, the rectangle PQRS have maximum perimeter.

So length of sides

PS = QR = 2\left( {{\pi \over 4} - {\pi \over 6}} \right) = {\pi \over 6}

and

PQ = RS = 2\sin \left( {{\pi \over 3}} \right) = \sqrt 3

\therefore

Required area =

{\pi \over 6} \times \sqrt 3 = {\pi \over {2\sqrt 3 }}

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