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 Question:
This is a minimization problem. We have a fixed total length (perimeter) and want to minimize the total area of two shapes.
Step 2: Key Formula or Approach:
Let \( x \) be the length used for the square and \( (8 - x) \) be the length for the circle.
Side of square \( a = \frac{x}{4} \).
Circumference of circle \( 2\pi r = 8 - x \Rightarrow r = \frac{8 - x}{2\pi} \).
Total Area \( A = a^2 + \pi r^2 \).
Step 3: Detailed Explanation:
\[ A(x) = \left( \frac{x}{4} \right)^2 + \pi \left( \frac{8 - x}{2\pi} \right)^2 = \frac{x^2}{16} + \frac{(8 - x)^2}{4\pi} \]
To find the minimum, differentiate \( A(x) \) with respect to \( x \) and set to zero:
\[ A'(x) = \frac{2x}{16} + \frac{2(8 - x)(-1)}{4\pi} = 0 \]
\[ \frac{x}{8} - \frac{8 - x}{2\pi} = 0 \Rightarrow \frac{\pi x - 4(8 - x)}{8\pi} = 0 \]
\[ \pi x - 32 + 4x = 0 \Rightarrow x(\pi + 4) = 32 \Rightarrow x = \frac{32}{\pi + 4} \]
Now, substitute this \( x \) back into the area formula:
\[ A = \frac{1}{16} \left( \frac{32}{\pi + 4} \right)^2 + \frac{1}{4\pi} \left( 8 - \frac{32}{\pi + 4} \right)^2 \]
\[ A = \frac{1}{16} \cdot \frac{1024}{(\pi + 4)^2} + \frac{1}{4\pi} \left( \frac{8\pi + 32 - 32}{\pi + 4} \right)^2 \]
\[ A = \frac{64}{(\pi + 4)^2} + \frac{1}{4\pi} \cdot \frac{64\pi^2}{(\pi + 4)^2} \]
\[ A = \frac{64}{(\pi + 4)^2} + \frac{16\pi}{(\pi + 4)^2} = \frac{16(4 + \pi)}{(\pi + 4)^2} = \frac{16}{\pi + 4} \]
Step 4: Final Answer:
The least sum of the areas is \( \frac{16}{\pi + 4} \).