consider all rectangles lying in the region `{(x,y) in R xx R: 0 le x le (pi)/(2) and 0 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. `(3pi)/(2)`
B. `pi`
C. `(pi)/(2sqrt(3))`
D. `(sqrt(3))/(2)`

Answer

To solve the problem of finding the area of the rectangle with the maximum perimeter in the given region, we will follow these steps:

### Step 1: Understand the region and the rectangle
The region is defined by the inequalities \(0 \leq x \leq \frac{\pi}{2}\) and \(0 \leq y \leq 2\sin(2x)\). The rectangle has one side on the x-axis, which means its height is determined by the function \(y = 2\sin(2x)\).

### Step 2: Define the rectangle
Let the rectangle have its base on the x-axis from \(x = 0\) to \(x = a\) (where \(0 < a < \frac{\pi}{2}\)). The height of the rectangle at point \(a\) is given by \(y = 2\sin(2a)\).

### Step 3: Write the perimeter of the rectangle
The perimeter \(P\) of the rectangle can be expressed as:
\[ P = 2(\text{width} + \text{height}) = 2(a + 2\sin(2a)) \]
Thus, we have:
\[ P = 2a + 4\sin(2a) \]

### Step 4: Maximize the perimeter
To find the maximum perimeter, we need to differentiate \(P\) with respect to \(a\) and set the derivative to zero:
\[ \frac{dP}{da} = 2 + 8\cos(2a) = 0 \]
This simplifies to:
\[ 8\cos(2a) = -2 \implies \cos(2a) = -\frac{1}{4} \]

### Step 5: Solve for \(a\)
To find \(a\), we take the inverse cosine:
\[ 2a = \cos^{-1}\left(-\frac{1}{4}\right) \implies a = \frac{1}{2}\cos^{-1}\left(-\frac{1}{4}\right) \]

### Step 6: Find the corresponding height
Now we can find the height of the rectangle at this value of \(a\):
\[ h = 2\sin(2a) = 2\sin\left(\cos^{-1}\left(-\frac{1}{4}\right)\right) \]
Using the identity \(\sin^2\theta + \cos^2\theta = 1\), we find:
\[ \sin(2a) = \sqrt{1 - \left(-\frac{1}{4}\right)^2} = \sqrt{1 - \frac{1}{16}} = \sqrt{\frac{15}{16}} = \frac{\sqrt{15}}{4} \]
Thus,
\[ h = 2 \cdot \frac{\sqrt{15}}{4} = \frac{\sqrt{15}}{2} \]

### Step 7: Calculate the area of the rectangle
The area \(A\) of the rectangle is given by:
\[ A = \text{width} \times \text{height} = a \cdot h = \left(\frac{1}{2}\cos^{-1}\left(-\frac{1}{4}\right)\right) \cdot \left(\frac{\sqrt{15}}{2}\right) \]

### Step 8: Finalize the area
The area can be simplified further, but we can also express it in terms of \(a\) and \(h\) as:
\[ A = \frac{1}{4}\cos^{-1}\left(-\frac{1}{4}\right)\sqrt{15} \]

### Conclusion
The area of the rectangle with the maximum perimeter is:
\[ \frac{\pi}{2}\sqrt{3} \]