To solve this problem, let's break it down step by step:

Step 1: Define the variables

Let:

• $l$ = length of the rectangle
• $w$ = width of the rectangle

The perimeter $P$ of a rectangle is given by: $P = 2l + 2w$ For this problem, we are given that $P = 120$ meters, so: $2l + 2w = 120$ Simplifying this equation: $l + w = 60$ Thus, $w = 60 - l$.

Step 2: Find the area

The area $A$ of a rectangle is given by: $A = l \times w$ Substitute $w = 60 - l$ into the area equation: $A = l \times (60 - l) = 60l - l^2$ This is a quadratic equation in terms of $l$, which we can maximize by finding its vertex.

Step 3: Maximize the area

To maximize the quadratic function $A = 60l - l^2$, we find the vertex. For a quadratic equation of the form $ax^2 + bx + c$, the maximum (or minimum) occurs at $x = -\frac{b}{2a}$.

Here, $a = -1$ and $b = 60$, so the length that maximizes the area is: $l = -\frac{60}{2(-1)} = 30$ Since $l = 30$, and from $l + w = 60$, we find that $w = 60 - 30 = 30$.

Step 4: Result for maximum area

When $l = 30$ meters and $w = 30$ meters, the area is: $A = l \times w = 30 \times 30 = 900 \text{ square meters}$ Thus, the rectangle is a square with side lengths of 30 meters, and the maximum possible area is 900 square meters.

General Case: When the perimeter is $L$ units

For a general perimeter $L$, the equation becomes: $l + w = \frac{L}{2}$ So the area becomes: $A = l \times \left( \frac{L}{2} - l \right) = l \times \frac{L}{2} - l^2$ Maximizing this quadratic function, the length that gives the maximum area is: $l = \frac{L}{4}$ Thus, $w = \frac{L}{4}$, and the maximum area occurs when the rectangle is a square with side lengths of $\frac{L}{4}$.

The maximum area is: $A = \left( \frac{L}{4} \right)^2 = \frac{L^2}{16}$

Summary:

• For a perimeter of 120 meters, the maximum area is 900 square meters when the rectangle is a square with side lengths of 30 meters.
• For a general perimeter $L$, the maximum area is $\frac{L^2}{16}$ when the rectangle is a square with side lengths $\frac{L}{4}$.

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Do you want further details or have any questions?

Here are 5 relative questions:

1. How do we derive the formula for the maximum area of any given perimeter?
2. Why does the maximum area always occur when the rectangle is a square?
3. Can you explain the relationship between the perimeter and area in more detail?
4. How would this problem change if the rectangle were constrained to different shapes?
5. What would happen to the area if one dimension were fixed?

Tip: When dealing with quadratic equations, always remember that the maximum or minimum value occurs at the vertex, which is crucial for optimization problems!