Question 59 of 82advanced🔧 ApplyNumerical5 marks

A wire of length 28 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the length of the two pieces so that the combined area of the square and the circle is minimum?

Correct Answer

Let the length of the piece used to make the square be x metres. Then the length used for the circle is (28 – x) metres. The square has side x/4, so its area is (x/4)^2 = x2x^{2}/16. The circle has circumference 28 – x = 2πr, giving radius r = (28 – x)/(2π), and area πr2r^{2} = (28 – x)^2/(4π). Hence the combined area is A(x) = x2x^{2}/16 + (28 – x)^2/(4π), where 0 < x < 28.

To find the minimum, differentiate A(x) with respect to x: A'(x) = x/8 – (28 – x)/(2π). Set A'(x) = 0 to obtain the critical point: x/8 = (28 – x)/(2π). Cross-multiplying gives 2πx = 8(28 – x), leading to 2πx = 224 – 8x, so x(2π + 8) = 224, and x = 224/(2π + 8) = 112/(π + 4).

The second derivative is A''(x) = 1/8 + 1/(2π), which is positive for all x. Therefore, the critical point yields a minimum of the area.

Thus, the length for the square should be 112/(π + 4) m and the length for the circle should be 28 – 112/(π + 4) = 28π/(π + 4) m. These are the required lengths.

Exercise: EXERCISE 6.3 | Q: 22 | (Page 30)
For More Understanding

Explanation

This problem is solved by forming a function for the total area in terms of one variable, differentiating, and using the second derivative test to confirm a minimum. The method mirrors similar optimization examples in the textbook, such as Example 24 (minimizing sum of squares) and the box volume problem (using second derivative to find maxima). The key steps are expressing the areas of the square and circle from the given wire lengths, combining them, and applying calculus.

Solution Steps

  1. Step 1: Let x be the length used for the square; then the circle uses (28 – x).

  2. Step 2: Express areas: square area = x2x^{2}/16; circle area = (28 – x)^2/(4π).

  3. Step 3: Form total area function A(x) = x2x^{2}/16 + (28 – x)^2/(4π).

  4. Step 4: Differentiate: A'(x) = x/8 – (28 – x)/(2π).

  5. Step 5: Set A'(x)=0 and solve: x = 112/(π+4).

  6. Step 6: Compute A''(x)=1/8+1/(2π)>0, confirming minimum.

  7. Step 7: The other length is 28 – x = 28π/(π+4).

PreviousNext