1. **State the problem:** We want to find the dimensions of a rectangular field that maximize the enclosed area given a total fencing cost of 700. 2. **Define variables:** Let $x$ be the length of the horizontal sides (top and bottom) and $y$ be the length of the vertical sides. 3. **Write the cost equation:** The cost of fencing the vertical sides is $10$ per foot, so total vertical cost is $2 \times 10y = 20y$. The bottom costs $2$ per foot, so bottom cost is $2x$. The top costs $7$ per foot, so top cost is $7x$. Total cost: $$20y + 2x + 7x = 20y + 9x = 700$$ 4. **Express $y$ in terms of $x$:** $$20y = 700 - 9x$$ $$y = \frac{700 - 9x}{20}$$ 5. **Write the area function:** $$A = x \times y = x \times \frac{700 - 9x}{20} = \frac{700x - 9x^2}{20}$$ 6. **Maximize the area:** Take the derivative of $A$ with respect to $x$ and set it to zero. $$A'(x) = \frac{700 - 18x}{20} = 0$$ 7. **Solve for $x$:** $$700 - 18x = 0$$ $$18x = 700$$ $$x = \frac{700}{18} = \frac{350}{9} \approx 38.89$$ 8. **Find $y$:** $$y = \frac{700 - 9 \times \frac{350}{9}}{20} = \frac{700 - 350}{20} = \frac{350}{20} = 17.5$$ 9. **Check the second derivative:** $$A''(x) = \frac{-18}{20} = -0.9 < 0$$ This confirms a maximum. **Final answer:** The dimensions that maximize the area are approximately $$x = 38.89 \text{ ft (horizontal length)}$$ $$y = 17.5 \text{ ft (vertical length)}$$