Let’s solve this problem step by step.

Problem Analysis:

You are tasked with minimizing the total fencing required for a rectangular plot of land with an area of 1500 m². The plot is divided into two equal portions using an additional fence parallel to two sides.

Step 1: Represent the problem with variables.

• Let the dimensions of the rectangular plot be $x$ (length) and $y$ (width).
• Since the area of the plot is given, we know: $x \cdot y = 1500$

To divide the plot into two equal portions, an additional fence runs parallel to the width ($y$), so the total length of fencing required is: $F = 2x + 3y$ (2 lengths and 3 widths — 1 width for the dividing fence).

Step 2: Express $F$ as a function of one variable.

From $x \cdot y = 1500$, solve for $y$: $y = \frac{1500}{x}$

Substitute $y$ into $F$: $F(x) = 2x + 3 \cdot \frac{1500}{x}$ $F(x) = 2x + \frac{4500}{x}$

Step 3: Minimize $F(x)$.

To minimize $F(x)$, find its derivative and set it equal to zero: $F'(x) = 2 - \frac{4500}{x^2}$ Set $F'(x) = 0$: $2 = \frac{4500}{x^2}$ $x^2 = \frac{4500}{2} = 2250$ $x = \sqrt{2250} \approx 47.43 \, \text{m}$

Step 4: Find $y$ and verify the solution.

Substitute $x \approx 47.43$ into $y = \frac{1500}{x}$: $y = \frac{1500}{47.43} \approx 31.63 \, \text{m}$

Thus, the dimensions of the plot are approximately: $x \approx 47.43 \, \text{m}, \quad y \approx 31.63 \, \text{m}.$

Step 5: Verify fencing is minimized.

The second derivative of $F(x)$ is: $F''(x) = \frac{9000}{x^3}$ For $x > 0$, $F''(x) > 0$, confirming that $F(x)$ has a minimum at $x \approx 47.43$.

Final Answer:

The dimensions of the rectangular plot that minimize the fencing required are: $\boxed{x \approx 47.43 \, \text{m}, \, y \approx 31.63 \, \text{m}}$

Would you like further explanation or details? Here are 5 related questions to expand your understanding:

1. What is the physical significance of the additional fence in this problem?
2. How would the solution change if the plot area was larger or smaller?
3. Can we solve the problem geometrically instead of using calculus?
4. What are the practical implications of minimizing fencing in real-world scenarios?
5. How does the second derivative test confirm a minimum?

Tip: Always verify your work by checking if the solution satisfies the constraints of the problem!