Question

A wire 40 metre in length is to be cut into two pieces. One piece is formed into a square and the other piece into a circle. The lengths of the two pieces so that the combined area of the square and the circle is minimum, are respectively

• A. \(\frac{160}{π+4},\frac{40π}{π+4}\)
• B. \(\frac{π+4}{160},\frac{40π}{π+4}\)
• C. \(\frac{160}{π+4},\frac{π+4}{40π}\)
• D. \(30,10\)

Hint:

Let one piece be x metres, write the total area in x, then set the derivative to zero.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We have a wire of length 40 m cut into two pieces. One piece makes a square and the other makes a circle. We need the cut that makes the total area smallest, using calculus.

Step 2: Set up the area.
Let the square use \(x\) metres. Then its side is \(x/4\) and its area is \(x^2/16\). The circle uses \(40 - x\) metres, so its radius is \(r = \dfrac{40 - x}{2\pi}\) and its area is \(\pi r^2 = \dfrac{(40-x)^2}{4\pi}\).
\[ A(x) = \frac{x^2}{16} + \frac{(40 - x)^2}{4\pi} \]

Step 3: Find the critical point.
\[ A'(x) = \frac{x}{8} - \frac{40 - x}{2\pi} = 0 \Rightarrow \pi x = 4(40 - x) \Rightarrow x = \frac{160}{\pi + 4} \]

Step 4: Check it is a minimum.
\(A''(x) = \dfrac{1}{8} + \dfrac{1}{2\pi} > 0\), so the area is minimum at this \(x\).

Step 5: Find the other piece.
\[ 40 - x = 40 - \frac{160}{\pi+4} = \frac{40\pi}{\pi + 4} \]

Step 6: Check the options.
Option (A) gives \(\dfrac{160}{\pi+4}\) for the square and \(\dfrac{40\pi}{\pi+4}\) for the circle. Options (B) and (C) have lengths of the wrong form. Option (D) does not add up to the minimising split.

Final Answer:
The pieces are \(\dfrac{160}{\pi+4}\) m for the square and \(\dfrac{40\pi}{\pi+4}\) m for the circle, option (A). \[ \boxed{\frac{160}{\pi+4},\ \frac{40\pi}{\pi+4}} \]