A wire of length 2 units is cut into two parts which are bent respectively to form a square of side = x units and a cricle of radius = r units. If the sum of areas of the square and the circle so formed is minimum, then

Options

A. `2x=(pi+4)r`
B. `(4-pi)x=pir`
C. x=2r
D. 2x=r

Answer

To solve the problem of minimizing the sum of the areas of a square and a circle formed from a wire of length 2 units, we can follow these steps:

### Step 1: Define Variables
Let:
- \( x \) = side length of the square
- \( r \) = radius of the circle

### Step 2: Express the Length of the Wire
The total length of the wire is 2 units. If we use \( a \) units of wire to form the square, then the remaining length for the circle will be \( 2 - a \).

### Step 3: Relate the Length to the Square and Circle
The perimeter of the square is given by:
\[ 4x = a \implies x = \frac{a}{4} \]
The circumference of the circle is given by:
\[ 2\pi r = 2 - a \implies r = \frac{2 - a}{2\pi} \]

### Step 4: Calculate Areas
The area of the square \( A_s \) is:
\[ A_s = x^2 = \left(\frac{a}{4}\right)^2 = \frac{a^2}{16} \]
The area of the circle \( A_c \) is:
\[ A_c = \pi r^2 = \pi \left(\frac{2 - a}{2\pi}\right)^2 = \frac{(2 - a)^2}{4\pi} \]

### Step 5: Sum of Areas
The total area \( A \) is:
\[ A = A_s + A_c = \frac{a^2}{16} + \frac{(2 - a)^2}{4\pi} \]

### Step 6: Simplify the Expression
Expanding the area of the circle:
\[ A = \frac{a^2}{16} + \frac{(4 - 4a + a^2)}{4\pi} = \frac{a^2}{16} + \frac{4}{4\pi} - \frac{4a}{4\pi} + \frac{a^2}{4\pi} \]
Combining like terms:
\[ A = \left(\frac{1}{16} + \frac{1}{4\pi}\right)a^2 - \frac{1}{\pi}a + \frac{1}{\pi} \]

### Step 7: Differentiate and Find Critical Points
To minimize \( A \), we take the derivative \( A' \) and set it to zero:
\[ A' = 2\left(\frac{1}{16} + \frac{1}{4\pi}\right)a - \frac{4}{\pi} \]
Setting \( A' = 0 \):
\[ 2\left(\frac{1}{16} + \frac{1}{4\pi}\right)a = \frac{4}{\pi} \]
Solving for \( a \):
\[ a = \frac{4/\pi}{2\left(\frac{1}{16} + \frac{1}{4\pi}\right)} = \frac{2/\pi}{\frac{1}{16} + \frac{1}{4\pi}} \]

### Step 8: Solve for \( x \) and \( r \)
Using the value of \( a \) found, we can calculate \( x \) and \( r \):
\[ x = \frac{a}{4}, \quad r = \frac{2 - a}{2\pi} \]

### Final Step: Conclusion
Thus, we have derived the relationship between \( x \) and \( r \) in terms of the length of the wire used for each shape.