1. **Problem statement:** A 12 cm wire is cut into two parts. One part forms a square, the other forms a circle. We want to find the radius of the circle that minimizes the sum of their areas. 2. **Define variables:** Let $x$ be the length of the wire used for the square, so the remaining length $12 - x$ is used for the circle. 3. **Formulas:** - Perimeter of square = $x$, so each side = $\frac{x}{4}$. - Area of square = $\left(\frac{x}{4}\right)^2 = \frac{x^2}{16}$. - Circumference of circle = $12 - x$, so radius $r = \frac{12 - x}{2\pi}$. - Area of circle = $\pi r^2 = \pi \left(\frac{12 - x}{2\pi}\right)^2 = \frac{(12 - x)^2}{4\pi}$. 4. **Sum of areas:** $$A(x) = \frac{x^2}{16} + \frac{(12 - x)^2}{4\pi}$$ 5. **Minimize $A(x)$:** Differentiate and set derivative to zero. $$A'(x) = \frac{2x}{16} + \frac{2(12 - x)(-1)}{4\pi} = \frac{x}{8} - \frac{12 - x}{2\pi}$$ Set $A'(x) = 0$: $$\frac{x}{8} = \frac{12 - x}{2\pi}$$ 6. **Solve for $x$:** Multiply both sides by $8 \cdot 2\pi = 16\pi$: $$16\pi \cdot \frac{x}{8} = 16\pi \cdot \frac{12 - x}{2\pi}$$ $$2\pi x = 8(12 - x)$$ $$2\pi x = 96 - 8x$$ $$2\pi x + 8x = 96$$ $$x(2\pi + 8) = 96$$ $$x = \frac{96}{2\pi + 8}$$ 7. **Calculate radius $r$:** $$r = \frac{12 - x}{2\pi} = \frac{12 - \frac{96}{2\pi + 8}}{2\pi} = \frac{12(2\pi + 8) - 96}{(2\pi + 8)2\pi} = \frac{24\pi + 96 - 96}{(2\pi + 8)2\pi} = \frac{24\pi}{(2\pi + 8)2\pi} = \frac{24}{2(2\pi + 8)} = \frac{12}{2\pi + 8}$$ **Final answer:** $$\boxed{r = \frac{12}{2\pi + 8}}$$ This radius minimizes the sum of the areas of the square and the circle formed from the wire.