Optimization — Question 6

Question 6 - Solution

Let: $$x = \text{length used for the square}, \quad 0 < x < 100 \Rightarrow 100 - x = \text{length used for the circle}$$

Step 1: Area Expressions

Square: $$\text{Perimeter} = x \Rightarrow \text{Side} = \frac{x}{4},\quad A_{\text{square}} = \left( \frac{x}{4} \right)^2 = \frac{x^2}{16}$$

Circle: $$\text{Circumference} = 100 - x = 2\pi r \Rightarrow r = \frac{100 - x}{2\pi} \Rightarrow A_{\text{circle}} = \pi r^2 = \pi \left( \frac{100 - x}{2\pi} \right)^2 = \frac{(100 - x)^2}{4\pi}$$

Total Area: $$A(x) = \frac{x^2}{16} + \frac{(100 - x)^2}{4\pi}$$

Step 2: Minimize Area

Differentiate: $$A'(x) = \frac{1}{8}x - \frac{1}{2\pi}(100 - x)$$

Set $A'(x) = 0$: $$\frac{1}{8}x = \frac{1}{2\pi}(100 - x) \Rightarrow \pi x = 4(100 - x) \Rightarrow \pi x = 400 - 4x \Rightarrow x(\pi + 4) = 400 \Rightarrow x = \frac{400}{\pi + 4}$$

Step 3: Approximate and Justify Minimum

$$x \approx \frac{400}{3.1416 + 4} = \frac{400}{7.1416} \approx 56.0$$

$$\boxed{x \approx 56.0 \text{ cm for the square}},\quad \boxed{44.0 \text{ cm for the circle}}$$

Second derivative: $$A''(x) = \frac{1}{8} + \frac{1}{2\pi} > 0 \Rightarrow \text{This critical point is a minimum.}$$

$$\boxed{\text{This wire division minimizes the total area.}}$$