Question

A wire of length 28m 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 circle is minimized?

Hint:

In optimization problems involving minimizing or maximizing an area, always define the objective function, use constraints to express variables, and solve the resulting equation to find critical points.

Answer 1

Step 1: Define the variables.
Let \( x \) be the length of the wire used for the square, and \( y \) be the length of the wire used for the circle. Since the total length of the wire is 28m, we have: \[ x + y = 28 \] Step 2: Express the areas.
- The area of the square is given by: \[ \text{Area of square} = \left( \frac{x}{4} \right)^2 = \frac{x^2}{16} \] - The circumference of the circle is \( y \), so the radius of the circle is \( \frac{y}{2\pi} \). The area of the circle is: \[ \text{Area of circle} = \pi \left( \frac{y}{2\pi} \right)^2 = \frac{y^2}{4\pi} \] Step 3: Define the objective function.
The total area is the sum of the areas of the square and the circle: \[ A(x, y) = \frac{x^2}{16} + \frac{y^2}{4\pi} \] Step 4: Substitute \( y = 28 - x \).
Since \( x + y = 28 \), we substitute \( y = 28 - x \) into the objective function: \[ A(x) = \frac{x^2}{16} + \frac{(28 - x)^2}{4\pi} \] Step 5: Minimize the objective function.
To minimize the area, we take the derivative of \( A(x) \) with respect to \( x \) and set it equal to zero: \[ \frac{dA}{dx} = \frac{2x}{16} - \frac{2(28 - x)}{4\pi} \] Simplify: \[ \frac{dA}{dx} = \frac{x}{8} - \frac{28 - x}{2\pi} \] Set \( \frac{dA}{dx} = 0 \) to find the critical points: \[ \frac{x}{8} = \frac{28 - x}{2\pi} \] Step 6: Solve for \( x \).
Cross-multiply to solve for \( x \): \[ x \cdot 2\pi = 8 \cdot (28 - x) \] \[ 2\pi x = 224 - 8x \] \[ x(2\pi + 8) = 224 \] \[ x = \frac{224}{2\pi + 8} \] Step 7: Conclusion.
Thus, the length of the wire to be used for the square is \( x = \frac{224}{2\pi + 8} \). The length of the wire to be used for the circle is \( y = 28 - x \).