Optimization — Question 5

Question 5 - Solution

Let: $$x = \text{length of wire used for the square}, \quad 0 < x < 100$$ Then: $$100 - x = \text{length of wire used for the triangle}$$

Step 1: Expressions for Area

Square:

Let $s = \frac{x}{4}$ be the side of the square. Then area of the square is: $$A_{\text{square}} = s^2 = \left(\frac{x}{4}\right)^2 = \frac{x^2}{16}$$

Triangle:

Let each side of the triangle be $t = \frac{100 - x}{3}$. Then area of equilateral triangle is: $$A_{\text{triangle}} = \frac{\sqrt{3}}{4}t^2 = \frac{\sqrt{3}}{4} \left( \frac{100 - x}{3} \right)^2 = \frac{\sqrt{3}}{36}(100 - x)^2$$

Total Area: $$A(x) = \frac{x^2}{16} + \frac{\sqrt{3}}{36}(100 - x)^2$$

Step 2: Minimize Total Area

Differentiate: $$A'(x) = \frac{1}{8}x - \frac{\sqrt{3}}{18}(100 - x)$$

Set $A'(x) = 0$: $$\frac{1}{8}x = \frac{\sqrt{3}}{18}(100 - x) \Rightarrow 18x = 8\sqrt{3}(100 - x) \Rightarrow 18x = 800\sqrt{3} - 8\sqrt{3}x \Rightarrow (18 + 8\sqrt{3})x = 800\sqrt{3} \Rightarrow x = \frac{800\sqrt{3}}{18 + 8\sqrt{3}}$$

Rationalizing denominator: $$x = \frac{800\sqrt{3}(18 - 8\sqrt{3})}{(18 + 8\sqrt{3})(18 - 8\sqrt{3})} = \frac{800\sqrt{3}(18 - 8\sqrt{3})}{324 - 192} = \frac{800\sqrt{3}(18 - 8\sqrt{3})}{132}$$

Approximate: $$x \approx 43.5 \Rightarrow \text{Use about } \boxed{43.5 \text{ cm for the square}},\quad \boxed{56.5 \text{ cm for the triangle}}$$

Step 3: Justification for Minimum

We compute the second derivative: $$A''(x) = \frac{1}{8} + \frac{\sqrt{3}}{18} > 0$$

Since $A''(x) > 0$, this critical point gives a local minimum.

$$\boxed{\text{The wire distribution minimizes total area.}}$$