Question

A wire of length 22 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into an equilateral triangle. Then, the length of the side of the equilateral triangle, so that the combined area of the square and the equilateral triangle is minimum, is

• A. \(\frac{22}{9+4\sqrt3}\)
• B. \(\frac{66}{9+4\sqrt3}\)
• C. \(\frac{22}{4+9\sqrt3}\)
• D. \(\frac{66}{4+9\sqrt3}\)

Answer 1

The Correct Option is B

To solve the given problem, we need to minimize the combined area of a square and an equilateral triangle formed by a wire of total length 22 meters.

1. Let the length of the piece used to form the square be \( x \). Thus, the remaining part of the wire is \( 22 - x \) meters, which will be used to form the equilateral triangle.
2. The perimeter of a square is four times the side of the square. Hence: \[ 4s = x \implies s = \frac{x}{4} \] The area of the square is given by: \[ A_{\text{square}} = s^2 = \left(\frac{x}{4}\right)^2 = \frac{x^2}{16} \]
3. The perimeter of an equilateral triangle is three times the length of its side. Hence: \[ 3t = 22 - x \implies t = \frac{22 - x}{3} \] The area of an equilateral triangle is given by: \[ A_{\text{triangle}} = \frac{\sqrt{3}}{4} t^2 = \frac{\sqrt{3}}{4} \left(\frac{22 - x}{3}\right)^2 \]
4. We need to minimize the total area \( A_{\text{total}} = A_{\text{square}} + A_{\text{triangle}} \): \[ A_{\text{total}} = \frac{x^2}{16} + \frac{\sqrt{3}}{4} \left(\frac{22-x}{3}\right)^2 \]
5. To find the value of \( x \) that minimizes \( A_{\text{total}} \), we take the derivative of \( A_{\text{total}} \) with respect to \( x \), and set it to zero:
   A'_{\text{total}} = \frac{x}{8} - \frac{\sqrt{3} \cdot 2(22-x)}{36} = 0
   \frac{x}{8} = \frac{\sqrt{3}(22-x)}{18}
6. Solving this equation, we find: \[ 9x = 4\sqrt{3}(22-x) \implies 9x = 88\sqrt{3} - 4\sqrt{3}x \implies 9x + 4\sqrt{3}x = 88\sqrt{3} \] \[ x(9 + 4\sqrt{3}) = 88\sqrt{3} \implies x = \frac{88\sqrt{3}}{9 + 4\sqrt{3}} \]
7. Now we need the length of the side of the equilateral triangle: \[ t = \frac{22 - x}{3} = \frac{22 - \frac{88\sqrt{3}}{9 + 4\sqrt{3}}}{3} \] Solving, we find: \[ t = \frac{66}{9 + 4\sqrt{3}} \]

Therefore, the length of the side of the equilateral triangle that minimizes the combined area is \( \frac{66}{9 + 4\sqrt{3}} \).