1. **Problem:** We want to enclose a rectangular field with 500 m of fencing, using one side as a building (no fence needed on that side). Find the dimensions that maximize the enclosed area. 2. **Formula and rules:** - Let the length parallel to the building be $x$ meters. - Let the width perpendicular to the building be $y$ meters. - Since one side is the building, fencing is needed for two widths and one length: total fencing used is $x + 2y = 500$. - The area $A$ to maximize is $A = x \times y$. 3. **Express $y$ in terms of $x$ using the fencing constraint:** $$x + 2y = 500 \implies 2y = 500 - x \implies y = \frac{500 - x}{2}$$ 4. **Write area $A$ as a function of $x$ only:** $$A(x) = x \times \frac{500 - x}{2} = \frac{500x - x^2}{2} = 250x - \frac{x^2}{2}$$ 5. **Maximize $A(x)$ by finding critical points:** - Differentiate $A(x)$ with respect to $x$: $$A'(x) = 250 - x$$ - Set derivative to zero to find critical points: $$250 - x = 0 \implies x = 250$$ 6. **Check the nature of critical point:** - Second derivative: $$A''(x) = -1 < 0$$ - Since $A''(250) < 0$, $x=250$ is a maximum. 7. **Find corresponding $y$ value:** $$y = \frac{500 - 250}{2} = \frac{250}{2} = 125$$ 8. **Answer:** - The dimensions that maximize the area are: $$\boxed{x = 250 \text{ m (length)}, \quad y = 125 \text{ m (width)}}$$ - Maximum area is: $$A = 250 \times 125 = 31250 \text{ m}^2$$