A box without a lid is constructed from a 28 inch by 28 inch piece of cardboard by cutting x in. squares from each corner and folding up the sides. A) Determine the volume of the box as a function of the variable x B) \approx imate the values of x that prod

A) To determine the volume of the box as a function of the variable x, we first need to determine the length, width, and height of the box in terms of x. If we cut x inches from each corner of the cardboard, the length and width of the base of the box will each be 28 - 2x inches. To form the sides of the box, we fold up the edges of the cardboard, which will have a height of x inches. Therefore, the length, width, and height of the box will be: Length = 28 - 2x inches Width = 28 - 2x inches Height = x inches The volume of the box is given by the formula: Volume = Length x Width x Height Substituting the expressions for length, width, and height, we get: Volume = (28 - 2x) x (28 - 2x) x x Volume = x(28 - 2x)^2 Therefore, the volume of the box as a function of the variable x is given by the expression x(28 - 2x)^2. B) To \approx imate the values of x that produce a volume of 1543.5, we need to solve the equation: x(28 - 2x)^2 = 1543.5 This is a cubic equation, which can be difficult to solve exactly. However, we can use a numerical method, such as Newton's method or the bisection method, to \approx imate the solutions. Using a graphing calculator or spreadsheet software, we can plot the function y = x(28 - 2x)^2 and find the x-value(s) where the graph intersects the horizontal line y = 1543.5. Alternatively, we can use trial and error to test different values of x and see which ones produce a volume close to 1543.5. After trying different values, we find that the largest reasonable answer is x ? 7.53 inches (rounded to two decimal places). This produces a volume of \approx imately 1543.47 cubic inches, which is very close to the desired volume. Note that the term "largest reasonable answer" implies that we should only consider values of x that produce a physically feasible box. In this case, we should only consider values of x that are less than half of the length and width of the cardboard (i.e., x < 14 inches), since cutting more than half would result in a box with negative dimensions or no box at all.