To find the dimensions of a rectangle that maximizes the area while having a fixed perimeter, we can use the relationship between perimeter, area, and the properties of rectangles.
1. **Understanding the Problem**: The perimeter (P) of a rectangle is given by the formula:
P = 2(l + w)
where l is the length and w is the width. We know the perimeter is 108 meters, so:
2(l + w) = 108
This simplifies to:
l + w = 54
2. **Expressing Area**: The area (A) of the rectangle is given by:
A = l * w
Using the equation for perimeter, we can express one variable in terms of the other. Let’s solve for w:
w = 54 – l
Now we can substitute this into the area formula:
A = l * (54 – l) = 54l – l^2
3. **Maximizing the Area**: To find the maximum area, we can either use calculus or recognize that this is a quadratic equation that opens downwards (since the coefficient of l^2 is negative). The maximum value occurs at the vertex of the parabola.
The vertex can be found using the formula:
l = -b/(2a)
where A = -l^2 + 54l, thus a = -1 and b = 54:
l = -54/(2 * -1) = 27
4. **Finding the Width**: Now, we can find w using w = 54 – l:
w = 54 – 27 = 27
5. **Conclusion**: The dimensions of the rectangle that maximize the area, given the perimeter of 108 meters, are a length of 27 meters and a width of 27 meters. This means the rectangle is actually a square!
In summary, the optimal dimensions are:
Length: 27m, Width: 27m