Finding the Biggest Area for a Rectangle with a Fixed Perimeter

When we want to figure out the biggest area of a rectangle that has a set perimeter, we can use some math techniques. This involves looking at how the length and width of the rectangle relate to its area.

Step 1: Understanding the Basics

Let's say the length of the rectangle is $l$ and the width is $w$. To find the perimeter $P$ of a rectangle, we can use this formula:

$$P = 2(l + w)$$

If we know what the perimeter is, we can rewrite the width like this:

$$w = \frac{P}{2} - l$$

Step 2: Area in One Variable

The area $A$ of the rectangle is calculated with:

$$A = l \cdot w$$

If we put our expression for $w$ into this formula, we get:

$$A = l \left(\frac{P}{2} - l\right)$$

When we expand this, it looks like this:

$$A = \frac{P}{2}l - l^2$$

Step 3: Recognizing the Quadratic Function

The area formula we have, $A = -l^2 + \frac{P}{2}l$, is a quadratic equation. It has a standard form that looks like this: $A = ax^2 + bx + c$, where:

• $a = -1$ (the number in front of $l^2$)
• $b = \frac{P}{2}$ (the number in front of $l$)
• $c = 0$ (the constant number)

Step 4: Finding the Highest Point

The biggest area for the rectangle shows up at a special point called the vertex of the parabola from our quadratic equation. We can find the $l$-coordinate of this point using:

$$l = -\frac{b}{2a}$$

By plugging in our values for $a$ and $b$, we get:

$$l = -\frac{\frac{P}{2}}{2 \times -1} = \frac{P}{4}$$

Step 5: Figuring out the Width

Now that we know the length $l$, we can find the width $w$ using the formula we made earlier:

$$w = \frac{P}{2} - l = \frac{P}{2} - \frac{P}{4} = \frac{P}{4}$$

Step 6: Putting it All Together

So, to get the maximum area, both the length and the width are equal. This means the rectangle is actually a square. The biggest area $A_{max}$ is:

$$A_{max} = l \cdot w = \frac{P}{4} \cdot \frac{P}{4} = \frac{P^2}{16}$$

Example Calculation

Let’s say the perimeter of the rectangle is $P = 20$ units. Here's how we find the maximum area:

1. Length: $l = \frac{20}{4} = 5$ units
2. Width: $w = 5$ units
3. Maximum Area:

$$A_{max} = \frac{20^2}{16} = \frac{400}{16} = 25 \text{ square units}$$

Quick Summary

To sum it up, when we have a fixed perimeter, using quadratic equations helps us find the size of a rectangle that gives the biggest area. We see that the rectangle that provides the largest area is a square. This shows how useful quadratic equations can be in real-life problems and why learning about them is important in math.