Minimizing total fencing for a divided rectangular garden

The total fencing consists of two sides of length $x$ (top and bottom) and three segments of length $y$ (left side, right side, and the middle divider), so $F = 2x + 3y$. The area constraint is $xy = 800$, so $y = \dfrac{800}{x}$. Substituting gives $F(x) = 2x + \dfrac{2400}{x}$, defined for $x>0$.

Differentiate: $F'(x) = 2 - \dfrac{2400}{x^2}$. Setting $F'(x)=0$ gives $x^2 = 1200$, so $x = \sqrt{1200} = 20\sqrt{3}$. Since $F''(x) = \dfrac{4800}{x^3} > 0$ for $x>0$, this critical point is a minimum. Then $y = \dfrac{800}{20\sqrt{3}} = \dfrac{40}{\sqrt{3}} = \dfrac{40\sqrt{3}}{3}$, and the minimum fencing is $F = 2(20\sqrt{3}) + 3\left(\dfrac{40\sqrt{3}}{3}\right) = 40\sqrt{3}+40\sqrt{3}=80\sqrt{3} \approx 138.56 \text{ m}$.