1. **Problem Statement:** Two roads A and B intersect perpendicularly. A diversion road C passes through a community located 4 km from road A and 6 km from road B. Find the smallest area bounded by roads A, B, and C. 2. **Understanding the problem:** Roads A and B are perpendicular, so they form a right angle. The diversion road C cuts across, forming a triangle with A and B. The community is at a point 4 km from A and 6 km from B, so the point lies at coordinates (4,6) if we consider A along the x-axis and B along the y-axis. 3. **Formula and approach:** The area of the triangle formed by roads A, B, and C is given by: $$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$$ Here, base and height correspond to the distances along roads A and B where road C intersects. 4. **Setting variables:** Let the diversion road C intersect road A at point $(x,0)$ and road B at point $(0,y)$. 5. **Equation of line C:** The line passing through $(x,0)$ and $(0,y)$ has equation: $$y = -\frac{y}{x}x + y$$ 6. **Condition for passing through community point (4,6):** Substitute $(4,6)$ into the line equation: $$6 = -\frac{y}{x} \times 4 + y$$ Simplify: $$6 = y - \frac{4y}{x} = y\left(1 - \frac{4}{x}\right)$$ 7. **Solve for y:** $$y = \frac{6}{1 - \frac{4}{x}} = \frac{6x}{x - 4}$$ 8. **Area expression:** $$A = \frac{1}{2} x y = \frac{1}{2} x \times \frac{6x}{x - 4} = \frac{3x^2}{x - 4}$$ 9. **Domain:** Since $y$ must be positive, $x > 4$. 10. **Minimize area:** Find derivative of $A$ with respect to $x$: $$A = \frac{3x^2}{x - 4}$$ Using quotient rule: $$A' = \frac{3(2x)(x - 4) - 3x^2(1)}{(x - 4)^2} = \frac{6x(x - 4) - 3x^2}{(x - 4)^2} = \frac{6x^2 - 24x - 3x^2}{(x - 4)^2} = \frac{3x^2 - 24x}{(x - 4)^2}$$ Set $A' = 0$: $$3x^2 - 24x = 0 \Rightarrow 3x(x - 8) = 0$$ So $x = 0$ or $x = 8$. Since $x > 4$, $x = 8$. 11. **Find corresponding y:** $$y = \frac{6 \times 8}{8 - 4} = \frac{48}{4} = 12$$ 12. **Minimum area:** $$A = \frac{1}{2} \times 8 \times 12 = 48$$ **Final answer:** \boxed{48 \text{ square kilometers}}