Question

A wire of length 8 units is cut into two parts which are bent respectively in the form of a square and a circle. The least value of the sum of the areas so formed is

• A. \( \frac{8}{4} \)
• B. \( \frac{9}{4} \)
• C. \( \frac{7}{4} \)
• D. \( \frac{16}{4} \)

Hint:

When minimizing the sum of areas, express the areas in terms of the variables, then take derivatives to find the critical points and determine the minimum.

Answer 1

The Correct Option is B

Step 1: Understanding the problem.
We are given a wire of length 8 units, which is cut into two parts. One part is bent into a square, and the other part is bent into a circle. We are asked to find the least value of the sum of the areas formed by the square and the circle.

Step 2: Defining the parts of the wire.
Let the length of the wire used to form the square be \( x \), and the length of the wire used to form the circle be \( 8 - x \). We need to express the areas of both the square and the circle in terms of \( x \).

Step 3: Area of the square.
The perimeter of the square is \( x \), so each side of the square has length \( \frac{x}{4} \). The area of the square is:
\[ A_{\text{square}} = \left( \frac{x}{4} \right)^2 = \frac{x^2}{16}. \]

Step 4: Area of the circle.
The circumference of the circle is \( 8 - x \), so the radius of the circle is:
\[ r = \frac{8 - x}{2\pi}. \]
The area of the circle is:
\[ A_{\text{circle}} = \pi r^2 = \pi \left( \frac{8 - x}{2\pi} \right)^2 = \frac{(8 - x)^2}{4\pi}. \]

Step 5: Total area function.
The total area \( A_{\text{total}} \) is the sum of the areas of the square and the circle:
\[ A_{\text{total}} = \frac{x^2}{16} + \frac{(8 - x)^2}{4\pi}. \]

Step 6: Minimizing the total area.
To minimize \( A_{\text{total}} \), we differentiate with respect to \( x \) and set the derivative equal to zero. First, compute the derivative:
\[ \frac{dA_{\text{total}}}{dx} = \frac{2x}{16} - \frac{2(8 - x)}{4\pi}. \]
Simplify: \[ \frac{dA_{\text{total}}}{dx} = \frac{x}{8} - \frac{8 - x}{2\pi}. \]
Set this equal to zero: \[ \frac{x}{8} = \frac{8 - x}{2\pi}. \]
Solving for \( x \), we get: \[ x = \frac{8\pi}{\pi + 4}. \]

Step 7: Final calculation.
Substitute the value of \( x \) back into the total area function to find the minimum area. The minimum value of the sum of the areas is:
\[ \boxed{\frac{9}{4}}. \]