JEE Advance - Mathematics (2020 - Paper 1 Offline - No. 6)
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
$$\left\{ {(x,y) \in R \times R:0\, \le \,x\, \le \,{\pi \over 2}} \right.$$ 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
$${{3\pi \over 2}}$$
$$\pi $$
$${\pi \over {2\sqrt 3 }}$$
$${{\pi \sqrt 3 } \over 2}$$
Explanation
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 $${\left. {{{{d^2}y} \over {d{x^2}}}} \right|_{x = {\pi \over 6}}} = {\left. { - 4\sin 2x} \right|_{x = {\pi \over 6}}}\, < \,0$$
$$ \therefore $$ 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 = $${\pi \over 6} \times \sqrt 3 = {\pi \over {2\sqrt 3 }}$$
$$\{ (x,y) \in R \times R:0\, \le \,x\, \le \,{\pi \over 2}$$
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 $${\left. {{{{d^2}y} \over {d{x^2}}}} \right|_{x = {\pi \over 6}}} = {\left. { - 4\sin 2x} \right|_{x = {\pi \over 6}}}\, < \,0$$
$$ \therefore $$ 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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