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The perimeter of a rectangle is the total distance around the outside of the rectangle. It is calculated by adding together twice the length and twice the width of the rectangle. This is expressed with the formula:\[ P = 2(l + w) \]where \( P \) represents the perimeter, \( l \) is the length, and \( w \) is the width. By knowing the perimeter, one can find either \( l \) or \( w \) if the other dimension is provided. In our problem, the perimeter is given as 64 cm. This allows us to rewrite the perimeter formula as:\[ 2(l + w) = 64 \]Simplifying, we get \( l + w = 32 \). This equation becomes crucial in determining the values that will maximize the rectangle's area.

Optimizing the area of a rectangle given a fixed perimeter is a common problem in calculus. The area \( A \) of a rectangle is found with the formula:\[ A = l \times w \]To find the maximum area, we need to look at possible configurations of \( l \) and \( w \). Given that \( l + w = 32 \), you can express \( w \) in terms of \( l \) as \( w = 32 - l \), then substitute into the area formula:\[ A = l \times (32 - l) \]This quadratic expression in terms of \( l \) allows us to apply calculus techniques or vertex analysis of a parabola to determine where the maximum area occurs.

The expression for the area \( A = l \times (32 - l) = 32l - l^2 \) is a quadratic function of \( l \). Quadratic functions have the general form:\[ ax^2 + bx + c \]In this case, our function is \(-l^2 + 32l\), where \( a = -1 \) and \( b = 32 \). The negative \( a \) value indicates that the parabola, when graphed, opens downward, which is evident because we are seeking a maximum point.Quadratic functions like this are particularly useful for determining extremes—maximums or minimums—within a given set of conditions. This is because they offer a clear, visual representation of the relationship between variables when plotted.

The vertex of the parabola represented by a quadratic function is the point at which either the maximum or minimum value of the function occurs. For a quadratic function \( ax^2 + bx + c \), the vertex \( x \) value can be found at:\[ x = -\frac{b}{2a} \]In our area maximization problem, substituting \( a = -1 \) and \( b = 32 \) into this formula gives:\[ l = -\frac{32}{2(-1)} = 16 \]Thus, the length that maximizes the area is 16 cm. Using the relation \( w = 32 - l \), the corresponding width is also 16 cm. Consequently, the rectangle becomes a square, which allows for the maximum possible area under the given perimeter constraint. Understanding the vertex helps to locate the optimal solution efficiently and effectively.