1. **State the problem:** We have a strip of tin 14 inches long, bent to form a trapezoidal gutter with one side bent up at a 45° angle. We want to find the width of the base that maximizes the carrying capacity (cross-sectional area). 2. **Define variables and setup:** Let the base width be $x$ inches. Let the vertical side height be $h$ inches. The side bent at 45° will have length $s$ inches. 3. **Perimeter constraint:** The total length of the tin strip is 14 inches, so: $$x + h + s = 14$$ 4. **Relate $h$ and $s$ using the 45° angle:** Since the side is bent at 45°, the vertical height $h$ and horizontal projection of $s$ are equal: $$h = s \sin 45^\circ = s \frac{\sqrt{2}}{2}$$ 5. **Express $s$ in terms of $h$:** $$s = \frac{h}{\sin 45^\circ} = h \sqrt{2}$$ 6. **Rewrite the perimeter constraint:** $$x + h + h \sqrt{2} = 14$$ $$x = 14 - h(1 + \sqrt{2})$$ 7. **Area of trapezoidal cross-section:** The gutter cross-section is trapezoidal with base $x$, one vertical side $h$, and the other side bent at 45° with horizontal projection $h$ (since $s \cos 45^\circ = h$). The top width is: $$x - h$$ Area $A$ is average of bases times height: $$A = \frac{(x + (x - h))}{2} \times h = \frac{(2x - h)}{2} h = h(x - \frac{h}{2})$$ 8. **Substitute $x$ from step 6:** $$A = h \left(14 - h(1 + \sqrt{2}) - \frac{h}{2} \right) = h \left(14 - h \left(1 + \sqrt{2} + \frac{1}{2} \right) \right)$$ 9. **Simplify the coefficient of $h$ inside parentheses:** $$1 + \sqrt{2} + \frac{1}{2} = \frac{3}{2} + \sqrt{2}$$ So, $$A = h \left(14 - h \left( \frac{3}{2} + \sqrt{2} \right) \right) = 14h - h^2 \left( \frac{3}{2} + \sqrt{2} \right)$$ 10. **Maximize area $A$ with respect to $h$:** Take derivative and set to zero: $$\frac{dA}{dh} = 14 - 2h \left( \frac{3}{2} + \sqrt{2} \right) = 0$$ 11. **Solve for $h$:** $$14 = 2h \left( \frac{3}{2} + \sqrt{2} \right)$$ $$h = \frac{14}{2 \left( \frac{3}{2} + \sqrt{2} \right)} = \frac{7}{\frac{3}{2} + \sqrt{2}}$$ 12. **Simplify denominator:** $$\frac{3}{2} + \sqrt{2} = \frac{3 + 2\sqrt{2}}{2}$$ So, $$h = \frac{7}{\frac{3 + 2\sqrt{2}}{2}} = \frac{7 \times 2}{3 + 2\sqrt{2}} = \frac{14}{3 + 2\sqrt{2}}$$ 13. **Rationalize denominator:** $$h = \frac{14}{3 + 2\sqrt{2}} \times \frac{3 - 2\sqrt{2}}{3 - 2\sqrt{2}} = \frac{14(3 - 2\sqrt{2})}{9 - 8} = 14(3 - 2\sqrt{2})$$ 14. **Calculate $h$ approximately:** $$3 - 2\sqrt{2} \approx 3 - 2 \times 1.414 = 3 - 2.828 = 0.172$$ $$h \approx 14 \times 0.172 = 2.408$$ 15. **Find $x$ from step 6:** $$x = 14 - h(1 + \sqrt{2}) = 14 - 2.408(1 + 1.414) = 14 - 2.408 \times 2.414 = 14 - 5.81 = 8.19$$ **Final answer:** The width of the base for greatest carrying capacity is approximately **8.19 inches**.