1. **Problem statement:** A window is shaped as a rectangle surmounted by an equilateral triangle. The total perimeter is 12 m. Find the rectangle dimensions that maximize the window's area. 2. **Step 1: Define variables and expressions.** Let the rectangle's width be $x$ and height be $y$. The equilateral triangle sits on top with side length $x$. 3. **Step 2: Perimeter expression.** The perimeter includes the rectangle's two heights and the base plus the two equal sides of the triangle (excluding the rectangle's top side since it's shared): $$P = 2y + 2x + x = 2y + 3x = 12$$ 4. **Step 3: Express $y$ in terms of $x$.** $$2y = 12 - 3x \implies y = \frac{12 - 3x}{2}$$ 5. **Step 4: Area expression.** Area $A$ is the rectangle area plus the equilateral triangle area: $$A = xy + \frac{\sqrt{3}}{4}x^2$$ Substitute $y$: $$A = x \cdot \frac{12 - 3x}{2} + \frac{\sqrt{3}}{4}x^2 = 6x - \frac{3}{2}x^2 + \frac{\sqrt{3}}{4}x^2$$ 6. **Step 5: Simplify area function.** $$A = 6x - \left(\frac{3}{2} - \frac{\sqrt{3}}{4}\right)x^2 = 6x - \left(\frac{6}{4} - \frac{\sqrt{3}}{4}\right)x^2 = 6x - \frac{6 - \sqrt{3}}{4}x^2$$ 7. **Step 6: Maximize area by differentiation.** $$\frac{dA}{dx} = 6 - 2 \cdot \frac{6 - \sqrt{3}}{4} x = 6 - \frac{6 - \sqrt{3}}{2} x$$ Set derivative to zero: $$6 - \frac{6 - \sqrt{3}}{2} x = 0 \implies x = \frac{12}{6 - \sqrt{3}}$$ 8. **Step 7: Rationalize denominator and approximate.** $$x = \frac{12}{6 - \sqrt{3}} \cdot \frac{6 + \sqrt{3}}{6 + \sqrt{3}} = \frac{12(6 + \sqrt{3})}{36 - 3} = \frac{12(6 + \sqrt{3})}{33} = \frac{72 + 12\sqrt{3}}{33}$$ Approximate: $$x \approx \frac{72 + 20.78}{33} = \frac{92.78}{33} \approx 2.81$$ 9. **Step 8: Find $y$.** $$y = \frac{12 - 3x}{2} = \frac{12 - 3(2.81)}{2} = \frac{12 - 8.43}{2} = \frac{3.57}{2} = 1.785$$ 10. **Answer:** The rectangle dimensions that maximize the area are approximately width $x = 2.81$ m and height $y = 1.79$ m.