\begin{tabular}{l} 16) Claire is building a rectangular frame to hold her stretched canvas painting. She has 14 feet of wood to build to the \\ frame. \\ Diagram: \\ a) Write an equation \( A(w) \) for the area of the stretched \\ canvas in terms of the width. \\ Calculator window: \( x:[\quad \) b) What is the maximum possible area for the painting? \\ \hline\end{tabular}

We start by letting the width of the rectangle be w (in feet). Since Claire has 14 feet of wood, and a rectangle has two widths and two lengths, the perimeter is given by   2(w + l) = 14 Solving for the length l:   w + l = 7  ⟹  l = 7 - w Now, the area A of the painting is the product of the width and the length:   A(w) = w(7 - w) = 7w - w² This is the equation for the area in terms of the width. To find the maximum possible area, notice that A(w) = 7w - w² is a quadratic function that opens downward (since the coefficient of w² is -1). The maximum occurs at the vertex of the parabola. For a quadratic in the form A(w) = aw² + bw + c, the vertex is found at   w = -b/(2a). Here, a = -1 and b = 7, so   w = -7/(2 * -1) = 7/2 = 3.5 feet. Substitute w = 3.5 back into the area function:   A(3.5) = 7(3.5) - (3.5)² = 24.5 - 12.25 = 12.25 square feet. Therefore: a) The area function is A(w) = 7w - w². b) The maximum possible area is 12.25 square feet.