Question

What is the maximum area of a rectangular enclosure with 1000m of fencing?

Answer 1

To find the maximum area of a rectangular enclosure using a fixed amount of fencing, we can use the following steps:

Let:
- \( L \) be the length of the rectangle,
- \( W \) be the width of the rectangle,
- The perimeter \( P \) is given by \( P = 2L + 2W \).

Given that the total amount of fencing is 1000 meters, we can set up the equation:

\[
2L + 2W = 1000
\]

Dividing the entire equation by 2, we have:

\[
L + W = 500
\]

Now, we want to express the area \( A \) of the rectangle, which is given by:

\[
A = L \times W
\]

We can express \( W \) in terms of \( L \):

\[
W = 500 - L
\]

Substituting this into the area formula gives:

\[
A = L(500 - L) = 500L - L^2
\]

This is a quadratic equation in \( L \), and it represents a downward-opening parabola. The maximum area occurs at the vertex of the parabola.

The vertex of a quadratic equation \( ax^2 + bx + c \) occurs at \( L = -\frac{b}{2a} \). Here, \( a = -1 \) and \( b = 500 \):

\[
L = -\frac{500}{2 \times -1} = \frac{500}{2} = 250
\]

Now substituting \( L = 250 \) back to find \( W \):

\[
W = 500 - 250 = 250
\]

Thus, the dimensions that maximize the area are \( L = 250 \) meters and \( W = 250 \) meters, which indicates the enclosure is a square.

Now, we can find the maximum area:

\[
A = L \times W = 250 \times 250 = 62500 \text{ square meters}
\]

Therefore, the maximum area of the rectangular enclosure with 1000 meters of fencing is:

\[
\boxed{62500} \text{ square meters}
\]