A rectangle is constructed such that the sum of its length and width is 54 . Let the length of the rectangle be \( x \). Form an expression for \( A \), the area of the rectangle, in terms of \( x \). What length of rectangle will result in the maximum possible area? Length \( =24 x-x^{2} \) What is the width of the rectangle with maximum area? Width = Enter your next step here

To find the width of the rectangle, we start by understanding that the sum of the length and width is 54. Since you defined the length as \( x \), we can express the width as \( 54 - x \). Now, substituting this into your area formula gives: \[ A = x(54 - x) = 54x - x^2 \] To maximize the area, we can take the derivative of \( A \), set it to zero, and solve for \( x \). Doing this, we find that the maximum area occurs when \( x = 27 \), which is half of 54. Now, plugging \( x = 27 \) back into the width expression: Width = \( 54 - 27 = 27 \). So, the width of the rectangle that results in the maximum area is 27. In this case, the rectangle is actually a square!