To find the dimensions of a rectangle that maximizes its area given a fixed perimeter of 100 meters, we can use the properties of rectangles and some basic algebra.
First, let's denote the length of the rectangle as L and the width as W. The formula for the perimeter (P) of a rectangle is:
P = 2(L + W)
Given that the perimeter is 100 meters, we have:
2(L + W) = 100
Dividing both sides by 2 gives us:
L + W = 50
Now, we can express the width in terms of the length:
W = 50 – L
Next, we need to find the area (A) of the rectangle, which is calculated as:
A = L × W
Substituting for W, we get:
A = L × (50 – L)
This simplifies to:
A = 50L – L²
To maximize the area, we need to find the critical points of the area function. We can do this by taking the derivative of the area function with respect to L and setting it to zero:
A’ = 50 – 2L = 0
Solving for L gives:
2L = 50
L = 25
Now, substituting L back into the equation for W:
W = 50 – 25 = 25
Thus, both the length and the width of the rectangle are 25 meters. This means that the rectangle with the largest possible area, given a perimeter of 100 meters, is actually a square with sides of 25 meters.
Finally, to find the maximum area, we can substitute 25 for L (or W) into the area formula:
A = 25 × 25 = 625 square meters
In summary, the dimensions of the rectangle that maximize the area while having a perimeter of 100 meters are:
Length: 25 meters
Width: 25 meters
This configuration results in a maximum area of 625 square meters.