Step 1: Understanding the Concept:
This is a standard optimization problem from calculus.
We are given a constraint: the total perimeter of a rectangle is fixed at 64 meters.
We need to maximize the area of this rectangle.
Usually, a rectangle has two dimensions (length and width), but we can use the perimeter constraint to express the area as a function of only one variable.
Once we have a function \( A(x) \), we find its maximum by taking the derivative and setting it to zero.
Step 2: Key Formula or Approach:
Let the length of the rectangle be \( l \) and the width be \( w \).
Perimeter \( P = 2(l + w) = 64 \implies l + w = 32 \).
From this, we can write \( w = 32 - l \).
Area \( A = l \cdot w \).
Substituting for \( w \): \( A(l) = l(32 - l) = 32l - l^2 \).
Step 3: Detailed Explanation:
To find the value of \( l \) that maximizes the area, we calculate the first derivative \( A'(l) \):
\[ \frac{dA}{dl} = 32 - 2l \]
Set the first derivative to zero to find the critical points:
\[ 32 - 2l = 0 \implies 2l = 32 \implies l = 16 \]
To ensure this critical point is a maximum, we look at the second derivative \( A''(l) \):
\[ \frac{d^2A}{dl^2} = -2 \]
Since the second derivative is negative (\(<0 \)), the area is indeed maximized when \( l = 16 \).
Now, calculate the width \( w \) using the perimeter constraint:
\[ w = 32 - l = 32 - 16 = 16 \]
Both dimensions are 16 m, which means the rectangle is a square.
Step 4: Final Answer:
The dimensions that maximize the area are 16 m and 16 m.
This matches option (A).