Consider all rectangles lying in the region \ (x, y) in R × R: 0 leq x leq 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

Mathematics · Application of Derivatives

JEE Advanced 2020 — Paper 1 — Question 34

Consider all rectangles lying in the region

$\left\{(x, y) \in R \times R: 0 \leq x \leq \frac{\pi}{2} \text { and } 0 \leq y \leq 2 \sin (2 x)\right\}$ and having one side on the x-axis.

The area of the rectangle which has the maximum perimeter among all such rectangles, is

0 \leq y \leq 2 \sin 2x, \quad 0 < x < \frac{\pi}{2}

\text{for rectangle } 2 \sin 2\alpha = 2 \sin 2\beta \implies \alpha + \beta = \frac{\pi}{2}

\text{Perimeter} = 2(\beta - \alpha) + 4 \sin 2\beta = 2\left( \beta - \left( \frac{\pi}{2} - \beta \right) \right) + 4 \sin 2\beta

= 2\left( \beta - \frac{\pi}{2} + \beta \right) + 4 \sin 2\beta

= 2\left( 2\beta - \frac{\pi}{2} \right) + 4 \sin 2\beta

= 4\beta - \pi + 4 \sin 2\beta

\frac{dp}{d\beta} = 4 + 8 \cos 2\beta = 0, \quad \cos 2\beta = -\frac{1}{2}, \quad \beta = \frac{\pi}{3}

\frac{d^2p}{d\beta^2} = -16 \sin 2\beta < 0 \quad \text{so } P \text{ is max}

\text{So, area of rectangle} = (\beta - \alpha) 2 \sin 2\beta = \left( \frac{\pi}{3} - \frac{\pi}{6} \right) \cdot 2 \sin \left( 2 \cdot \frac{\pi}{3} \right)

= \frac{\pi}{6} \times 2 \cdot \frac{\sqrt{3}}{2} = \frac{\pi \sqrt{3}}{6}

Answer key and solution verified before publishing.

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