A figure consists of a semi-circle with a rectangle on its diameter. Given that the parameter of the figure is 20 feei. In order that its area may be maximum, its length and breadth are equal.

Answer

To solve the problem step by step, we will analyze the figure consisting of a semicircle with a rectangle on its diameter. We need to find the dimensions of the rectangle such that the area of the entire figure is maximized, given that the perimeter is 20 feet.

### Step 1: Define the Variables
Let:
- \( R \) = radius of the semicircle
- \( L \) = length of the rectangle
- \( B \) = breadth of the rectangle

Since the rectangle is placed on the diameter of the semicircle, the breadth of the rectangle will be equal to the diameter of the semicircle, which is \( 2R \).

### Step 2: Write the Perimeter Equation
The perimeter \( P \) of the figure consists of the semicircle and the rectangle. The perimeter can be expressed as:
\[ P = \text{Length of semicircle} + \text{Length of rectangle} + \text{Length of rectangle} \]
This can be written as:
\[ P = \frac{1}{2} \pi R + 2L + 2R \]
Given that the perimeter is 20 feet, we have:
\[ \frac{1}{2} \pi R + 2L + 2R = 20 \]

### Step 3: Solve for \( L \)
Rearranging the equation to solve for \( L \):
\[ 2L = 20 - \frac{1}{2} \pi R - 2R \]
\[ L = 10 - \frac{1}{4} \pi R - R \]

### Step 4: Write the Area Function
The area \( A \) of the figure consists of the area of the semicircle and the area of the rectangle:
\[ A = \text{Area of semicircle} + \text{Area of rectangle} \]
This can be expressed as:
\[ A = \frac{1}{2} \pi R^2 + L \cdot B \]
Substituting \( B = 2R \) and the expression for \( L \):
\[ A = \frac{1}{2} \pi R^2 + (10 - \frac{1}{4} \pi R - R)(2R) \]
\[ A = \frac{1}{2} \pi R^2 + 20R - \frac{1}{2} \pi R^2 - 2R^2 \]
This simplifies to:
\[ A = 20R - 2R^2 \]

### Step 5: Maximize the Area
To find the maximum area, we take the derivative of \( A \) with respect to \( R \) and set it to zero:
\[ \frac{dA}{dR} = 20 - 4R \]
Setting the derivative equal to zero:
\[ 20 - 4R = 0 \implies R = 5 \]

### Step 6: Verify Maximum Area
To confirm that this is a maximum, we take the second derivative:
\[ \frac{d^2A}{dR^2} = -4 \]
Since the second derivative is negative, this indicates that the area function has a maximum at \( R = 5 \).

### Step 7: Find Dimensions
Now, substituting \( R = 5 \) back to find \( L \):
\[ L = 10 - \frac{1}{4} \pi (5) - 5 \]
Calculating \( L \):
\[ L = 10 - \frac{5\pi}{4} - 5 = 5 - \frac{5\pi}{4} \]
And the breadth \( B \):
\[ B = 2R = 10 \]

### Conclusion
Thus, the dimensions of the rectangle for maximum area are:
- Length \( L = 5 - \frac{5\pi}{4} \)
- Breadth \( B = 10 \)