Consider all rectangles lying in the region {(x, y) ∈ R × R : 0 ≤ x ≤ π/2 and 0 ≤ y ≤ 2sin(2x)} and

Q: Wait, why do we use $2\alpha$ as the width instead of just $x$?

Great question! Using $x$ as the variable for the right edge would make the width $x - (\pi/2 - x) = 2x - \pi/2$, which is messy. By using $\alpha$ as the distance from the center line $x = \pi/4$, we exploit the symmetry of the curve. This makes the width $2\alpha$ and the height $2\cos(2\alpha)$, which simplifies the calculus significantly.

Q: Why is the height $2\cos(2\alpha)$? I thought it was $2\sin(2x)$.

You are absolutely right that the height is determined by the curve $y = 2\sin(2x)$. When we plug in the x-coordinate of the edge, $x = \pi/4 - \alpha$, we get $y = 2\sin(2(\pi/4 - \alpha))$. Distributing the 2 gives $2\sin(\pi/2 - 2\alpha)$. Using the trigonometric identity $\sin(\pi/2 - \theta) = \cos(\theta)$, this transforms perfectly into $2\cos(2\alpha)$.

Q: The question asks for the Area, but we maximized the Perimeter. Did I miss a step?

This is a classic JEE trap! The problem asks us to find the area of the *specific* rectangle that has the maximum perimeter. We first find the $\alpha$ that maximizes the perimeter, and then we use that specific $\alpha$ to calculate the area. Always read the final sentence of the question carefully!

Question

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

Options:

(A) $\frac{3\pi}{2}$

(B) $\pi$

(C) $\frac{\pi}{2\sqrt{3}}$

(D) $\frac{\pi\sqrt{3}}{2}$

Sigma Percentile · Accepted Answer

### Analyzing the Geometry of the Hill

Imagine you are standing in a coordinate plane, looking at the curve defined by $y = 2\sin(2x)$ between $x = 0$ and $x = \pi/2$. This curve is a symmetric hill.

Our task is to fit a rectangle under this hill, with its base resting on the $x$-axis, and find the one that has the largest possible perimeter. Once we find the dimensions of this 'perfect' rectangle, we must calculate its area.

### The Power of Symmetry

The first step to solving any optimization problem is to simplify the geometry. The curve $y = 2\sin(2x)$ is perfectly symmetric about the line $x = \pi/4$.

It is intuitive that the rectangle should also be symmetric about that same line. Let us define the width of our rectangle as $2\alpha$. By centering the rectangle at $x = \pi/4$, the left edge sits at $x = \pi/4 - \alpha$ and the right edge sits at $x = \pi/4 + \alpha$.

### The Calculus of Optimization

The top corners of our rectangle must touch the curve. Plugging the left edge $x = \pi/4 - \alpha$ into the curve equation $y = 2\sin(2x)$, we get:

$$y = 2\sin\left(2\left(\frac{\pi}{4} - \alphaight)ight) = 2\sin\left(\frac{\pi}{2} - 2\alphaight) = 2\cos(2\alpha)$$

Now, we have the width $2\alpha$ and the height $2\cos(2\alpha)$. The perimeter $P$ is twice the sum of the width and height:

$$P(\alpha) = 2(2\alpha + 2\cos(2\alpha)) = 4(\alpha + \cos(2\alpha))$$

To find the maximum, we take the derivative with respect to $\alpha$:

$$\frac{dP}{d\alpha} = 4(1 - 2\sin(2\alpha))$$

Setting this to zero, we find $1 - 2\sin(2\alpha) = 0$, which implies $\sin(2\alpha) = 1/2$. This yields $2\alpha = \pi/6$, or $\alpha = \pi/12$.

A quick check with the second derivative, $\frac{d^2P}{d\alpha^2} = -8\cos(2\alpha)$, confirms that at $2\alpha = \pi/6$, the second derivative is negative, confirming a maximum.

### The Final Calculation

We have found the condition for the maximum perimeter, but we must now calculate the area $A$ of this specific rectangle. The area is defined as width times height:

$$A = (2\alpha)(2\cos(2\alpha))$$

Substituting our value $2\alpha = \pi/6$ into the expression, we get:

$$A = \left(\frac{\pi}{6}ight) 	imes 2\cos\left(\frac{\pi}{6}ight)$$

Since $\cos(\pi/6) = \sqrt{3}/2$, the final calculation becomes:

$$A = \frac{\pi}{6} 	imes 2 	imes \frac{\sqrt{3}}{2} = \frac{\pi\sqrt{3}}{6}$$

Through the power of symmetry and careful calculus, we have navigated the problem to find the **final area** of **$\frac{\pi\sqrt{3}}{6}$**.

Select Answer:

Visualized Solution

Visualize the Region

Identify Symmetry

Define Rectangle Dimensions

Determine Rectangle Height

Formulate Perimeter Function

Differentiate the Perimeter

Find the Critical Points

Verify Maximum via Second Derivative

Setup the Area Calculation

Compute the Final Area

The Sigma Insight: Maxima and Minima

Analyzing the Geometry of the Hill

The Power of Symmetry

The Calculus of Optimization

The Final Calculation

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