Question

A wire of length 36 m is cut into two pieces, one of the pieces is bent to form a square and the other is bent to form a circle. If the sum of the areas of the two figures is minimum, and the circumference of the circle is $k$ (meter), then $\left( \frac{4}{\pi} + 1 \right)k$ is equal to _________.

Hint:

For minimum total area of a circle and a square, the diameter of the circle is equal to the side of the square ($2r = a$).

Answer 1

Correct Answer: 36

Step 1: Understanding the Concept:
This is an optimization problem. Let the piece bent into a circle have length $L_1$ and the square have length $L_2$. We express the total area in terms of one variable and find the condition for its minimum using derivatives.
Step 2: Key Formula or Approach:
Total length $P = 2\pi r + 4a = 36$.
Total area $A = \pi r^2 + a^2$.
Step 2: Detailed Explanation:
From the perimeter equation: $4a = 36 - 2\pi r \implies a = 9 - \frac{\pi r}{2}$.
Substitute $a$ into the area equation:
\[ A(r) = \pi r^2 + \left( 9 - \frac{\pi r}{2} \right)^2 \]
Differentiate with respect to $r$ and set to zero for minimum:
\[ A'(r) = 2\pi r + 2\left( 9 - \frac{\pi r}{2} \right)\left( -\frac{\pi}{2} \right) = 0 \]
\[ 2\pi r - \pi \left( 9 - \frac{\pi r}{2} \right) = 0 \]
Divide by $\pi$:
\[ 2r - 9 + \frac{\pi r}{2} = 0 \implies 4r + \pi r = 18 \implies r = \frac{18}{\pi + 4} \]
The circumference of the circle $k = 2\pi r = \frac{36\pi}{\pi + 4}$.
The value to find is:
\[ \left( \frac{4}{\pi} + 1 \right) k = \left( \frac{4 + \pi}{\pi} \right) \cdot \frac{36\pi}{\pi + 4} = 36 \]
Step 3: Final Answer:
The value is 36.