If a field has perimeter P, what shape maximizes the area? Justify your reasoning.
Let's suppose the perimeter is 100 meters.
If it was 2 by 48 meters, the area would be would be 196 square meters
If it was 20 by 30 meters, the area would be 600 square meters.
If it was 25 by 25, the area would be 625 square meters.
But, what shapes are those measurements?
If a rectangle has four sides of equal length, what shape is it?
If all shapes have the same perimeter a circle will be the shape that would maximize the area.
To determine the shape that maximizes the area for a given perimeter, we can consider different shapes and compare their corresponding areas.
Let's start by analyzing a rectangle. Let the length of one side be L, and the width be W. The perimeter of a rectangle is calculated using the formula: P = 2L + 2W. Solving for L, we get L = (P - 2W)/2.
The area of a rectangle is given by A = L * W = ((P - 2W)/2) * W = 0.5PW - W^2/2. To maximize the area, we need to find the value of W that maximizes A.
Now, let's consider a square. In a square, all four sides are equal. So, the perimeter of a square is given by P = 4S, where S is the length of each side.
The area of a square is given by A = S^2.
We can compare the areas of the rectangle and the square by relating their respective perimeters:
2L + 2W = 4S (Equation 1)
From Equation 1, we can solve for W in terms of S:
L + W = 2S (Equation 2)
L = 2S - W
Substituting the value of L into the area equation for the rectangle, we get:
A = 0.5PW - W^2/2 = 0.5(2S - W)W - W^2/2 = SW - 0.5W^2
To find the shape that maximizes the area, let's differentiate A with respect to W and set it equal to zero, seeking the critical points:
dA/dW = S - W = 0
Solving for W, we find W = S.
This means that in order for the area A to be maximized, the width of the rectangle (W) must be equal to the side length of the square (S).
Therefore, the shape that maximizes the area for a given perimeter is a square. This conclusion can be supported mathematically by demonstrating that any other shape with the same perimeter will have a smaller area.