A window is the in the form of a reactangle, surmounted by a semi - circle. If the perimeter be 15 metres, find the dimesnions so that greatest possible amount of light may be admitted in order that its area may be maximum.

Answer

To solve the problem of maximizing the area of a window in the form of a rectangle surmounted by a semicircle with a given perimeter of 15 meters, we can follow these steps:

### Step 1: Define Variables
Let:
- \( x \) = length of the rectangle
- \( y \) = breadth of the rectangle
- The diameter of the semicircle is equal to the length of the rectangle, which is \( x \).
- The radius of the semicircle is \( r = \frac{x}{2} \).

### Step 2: Write the Perimeter Equation
The perimeter of the window consists of the length of the rectangle, the two sides of the rectangle, and the circumference of the semicircle. Thus, we can write:

\[ \text{Perimeter} = x + 2y + \text{Circumference of semicircle} \]

The circumference of the semicircle is given by:

\[ \text{Circumference} = \pi r = \pi \left(\frac{x}{2}\right) = \frac{\pi x}{2} \]

So, the perimeter equation becomes:

\[ x + 2y + \frac{\pi x}{2} = 15 \]

### Step 3: Solve for \( y \)
Rearranging the perimeter equation to solve for \( y \):

\[ 2y = 15 - x - \frac{\pi x}{2} \]

\[ y = \frac{15 - x - \frac{\pi x}{2}}{2} \]

### Step 4: Write the Area Equation
The area \( A \) of the window consists of the area of the rectangle and the area of the semicircle:

\[ A = \text{Area of rectangle} + \text{Area of semicircle} \]

The area of the rectangle is:

\[ \text{Area of rectangle} = x \cdot y \]

The area of the semicircle is:

\[ \text{Area of semicircle} = \frac{1}{2} \pi r^2 = \frac{1}{2} \pi \left(\frac{x}{2}\right)^2 = \frac{\pi x^2}{8} \]

Thus, the total area \( A \) is:

\[ A = xy + \frac{\pi x^2}{8} \]

Substituting \( y \) from the previous step:

\[ A = x \left(\frac{15 - x - \frac{\pi x}{2}}{2}\right) + \frac{\pi x^2}{8} \]

### Step 5: Simplify the Area Equation
Expanding the area equation:

\[ A = \frac{x(15 - x - \frac{\pi x}{2})}{2} + \frac{\pi x^2}{8} \]

\[ A = \frac{15x - x^2 - \frac{\pi x^2}{2}}{2} + \frac{\pi x^2}{8} \]

To combine the terms, we need a common denominator. The common denominator for \( 2 \) and \( 8 \) is \( 8 \):

\[ A = \frac{60x - 4x^2 - 2\pi x^2}{8} + \frac{\pi x^2}{8} \]

\[ A = \frac{60x - 4x^2 + \pi x^2 - 2\pi x^2}{8} \]

\[ A = \frac{60x - 4x^2 - \pi x^2}{8} \]

### Step 6: Differentiate the Area Function
To find the maximum area, we differentiate \( A \) with respect to \( x \):

\[ \frac{dA}{dx} = \frac{1}{8}(60 - 8x - \pi x) = 0 \]

Setting the derivative equal to zero:

\[ 60 - 8x - \pi x = 0 \]

\[ 8x + \pi x = 60 \]

\[ x(8 + \pi) = 60 \]

\[ x = \frac{60}{8 + \pi} \]

### Step 7: Find \( y \)
Now substituting \( x \) back into the equation for \( y \):

\[ y = \frac{15 - \frac{60}{8 + \pi} - \frac{\pi}{2} \cdot \frac{60}{8 + \pi}}{2} \]

### Step 8: Final Dimensions
Calculating \( y \):

1. Calculate \( \frac{\pi}{2} \cdot \frac{60}{8 + \pi} \).
2. Substitute back to find \( y \).

### Conclusion
The dimensions of the window that maximize the area while maintaining a perimeter of 15 meters are:

- Length \( x = \frac{60}{8 + \pi} \) meters
- Breadth \( y = \text{calculated value} \) meters