JEE Challenger
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MathHardAdv 2020 • Paper 1

Area of Rectangle with Maximum Perimeter Bounded by Sine Curve

For all rectangles situated within the region {(x,y)∈R×R:0≤x≤π2 and 0≤y≤2sin⁡(2x)}\{(x,y) \in \mathbb{R} \times \mathbb{R} : 0 \le x \le \frac{\pi}{2} \text{ and } 0 \le y \le 2\sin(2x)\} with one side lying along the xx-axis, determine the area of the rectangle that achieves the maximum perimeter.

Options

A

3π2\frac{3\pi}{2}

B

π\pi

C

π23\frac{\pi}{2\sqrt{3}}

Correct
D

π32\frac{\pi\sqrt{3}}{2}

Topics & Concepts

Application of DerivativesMaxima and Minima

Step-by-Step Solution

To find the area of the rectangle with the maximum perimeter lying within the region R={(x,y)∈R×R:0≤x≤π2 and 0≤y≤2sin⁡(2x)}\mathcal{R} = \left\{(x,y) \in \mathbb{R} \times \mathbb{R} : 0 \le x \le \frac{\pi}{2} \text{ and } 0 \le y \le 2\sin(2x)\right\} with one side lying on the xx-axis, we proceed as follows:

Let the vertices of the rectangle on the xx-axis be (x1,0)(x_1, 0) and (x2,0)(x_2, 0), where 0≤x1<x2≤π20 \le x_1 < x_2 \le \frac{\pi}{2}. The top two vertices lie on the curve y=2sin⁡(2x)y = 2\sin(2x), which means the heights at x1x_1 and x2x_2 must be equal: y=2sin⁡(2x1)=2sin⁡(2x2)y = 2\sin(2x_1) = 2\sin(2x_2)

Since 2x1,2x2∈[0,π]2x_1, 2x_2 \in [0, \pi] and the function sin⁡(θ)\sin(\theta) is symmetric about θ=π2\theta = \frac{\pi}{2}, we have: 2x2=π−2x1  ⟹  x2=π2−x12x_2 = \pi - 2x_1 \implies x_2 = \frac{\pi}{2} - x_1

Let x1=xx_1 = x, where x∈[0,π4]x \in \left[0, \frac{\pi}{4}\right]. Then:

  • Width of the rectangle, w=x2−x1=(π2−x)−x=π2−2xw = x_2 - x_1 = \left(\frac{\pi}{2} - x\right) - x = \frac{\pi}{2} - 2x
  • Height of the rectangle, h=2sin⁡(2x)h = 2\sin(2x)

Step 1: Maximize the Perimeter

The perimeter P(x)P(x) of the rectangle is given by: P(x)=2(w+h)=2(π2−2x+2sin⁡(2x))=π−4x+4sin⁡(2x)P(x) = 2(w + h) = 2\left(\frac{\pi}{2} - 2x + 2\sin(2x)\right) = \pi - 4x + 4\sin(2x)

To find the critical points, we take the derivative of P(x)P(x) with respect to xx: P′(x)=ddx(π−4x+4sin⁡(2x))=−4+8cos⁡(2x)P'(x) = \frac{d}{dx}\Big(\pi - 4x + 4\sin(2x)\Big) = -4 + 8\cos(2x)

Setting P′(x)=0P'(x) = 0: −4+8cos⁡(2x)=0  ⟹  cos⁡(2x)=12-4 + 8\cos(2x) = 0 \implies \cos(2x) = \frac{1}{2}

For x∈[0,π4]x \in \left[0, \frac{\pi}{4}\right], we have 2x∈[0,π2]2x \in \left[0, \frac{\pi}{2}\right]: 2x=π3  ⟹  x=π62x = \frac{\pi}{3} \implies x = \frac{\pi}{6}

To confirm that this gives a maximum, we check the second derivative: P′′(x)=−16sin⁡(2x)P''(x) = -16\sin(2x) At x=π6x = \frac{\pi}{6}: P′′(π6)=−16sin⁡(π3)=−83<0P''\left(\frac{\pi}{6}\right) = -16\sin\left(\frac{\pi}{3}\right) = -8\sqrt{3} < 0 Thus, x=π6x = \frac{\pi}{6} gives the maximum perimeter.

Step 2: Calculate the Area of the Rectangle

At x=π6x = \frac{\pi}{6}, the dimensions of the rectangle are:

  • Width: w=π2−2(π6)=π2−π3=π6w = \frac{\pi}{2} - 2\left(\frac{\pi}{6}\right) = \frac{\pi}{2} - \frac{\pi}{3} = \frac{\pi}{6}
  • Height: h=2sin⁡(2⋅π6)=2sin⁡(π3)=2(32)=3h = 2\sin\left(2 \cdot \frac{\pi}{6}\right) = 2\sin\left(\frac{\pi}{3}\right) = 2 \left(\frac{\sqrt{3}}{2}\right) = \sqrt{3}

Therefore, the area AA of the rectangle is: A=w⋅h=π6⋅3=π36=π23A = w \cdot h = \frac{\pi}{6} \cdot \sqrt{3} = \frac{\pi\sqrt{3}}{6} = \frac{\pi}{2\sqrt{3}}

Thus, the correct option is (C).

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