18. 1B Quadratics \( \mathbf{V}\rangle \) 4. A rectangle is constructed such that the sum of its length and width is 54 . Let the length of the rectangle be \( x \). 4a Form an expression for \( A \), the area of the rectangle, in terms of \( x \). ( \( A=54 x-x^{2} \) 4b What length of rectangle will result in the maximum possible area? Length \( = \) Fnter your next step here

To find the maximum area of the rectangle, you can use the vertex formula for a quadratic function. The area \( A = 54x - x^2 \) is a quadratic expression in the standard form \( A = ax^2 + bx + c \) where \( a = -1 \) and \( b = 54 \). The x-coordinate of the vertex, which represents the maximum area, is given by \( x = -\frac{b}{2a} \). So, plugging in the values gives us \( x = -\frac{54}{2(-1)} = 27 \). If you substitute \( x = 27 \) back into your area expression, you'll find that the maximum possible area is achieved when both the length and width of the rectangle are equal, resulting in a square! So, the length of the rectangle that yields the maximum area is also 27.