1. **Problem statement:** We have a rectangular field with an area of 600 m². We want to enclose it with fencing and also divide it into two equal halves with a fence of length $x$. We need to find the total length of fencing $f(x)$ in terms of $x$ and then find the minimum fencing required. 2. **Define variables:** Let $x$ be the length of the fence dividing the field into two equal halves. Since the field is divided into two equal halves, the dividing fence runs parallel to one side of the rectangle. 3. **Express dimensions:** Let the length of the field be $x$ (the dividing fence) and the width be $y$. The area is given by: $$xy = 600$$ 4. **Express $y$ in terms of $x$:** $$y = \frac{600}{x}$$ 5. **Total fencing length $f(x)$:** The fencing includes the perimeter plus the dividing fence. The perimeter is $2x + 2y$, and the dividing fence adds $x$ more. So, $$f(x) = 2x + 2y + x = 3x + 2y$$ Substitute $y$: $$f(x) = 3x + 2\frac{600}{x} = 3x + \frac{1200}{x}$$ 6. **Find minimum fencing:** To minimize $f(x)$, take the derivative and set it to zero: $$f'(x) = 3 - \frac{1200}{x^2}$$ Set $f'(x) = 0$: $$3 - \frac{1200}{x^2} = 0 \implies 3 = \frac{1200}{x^2} \implies x^2 = \frac{1200}{3} = 400$$ So, $$x = 20$$ 7. **Calculate minimum fencing length:** Substitute $x=20$ into $f(x)$: $$f(20) = 3(20) + \frac{1200}{20} = 60 + 60 = 120$$ **Final answers:** - a) $f(x) = 3x + \frac{1200}{x}$ - b) Minimum fencing required is $120$ meters when $x=20$ meters.