Question 3 of 30 A rectangular label with an area of 72 square metres is wrapped around a rectangular box 9 metres long. The label exactly covers the outside of the box, excluding the top and the bottom. If the top and the bottom of the box are both squares, what is the volume of the rectangular box?

To solve this problem, first, we need to find the dimensions of the box. The label is wrapped around the sides of the box, which has a length of 9 metres and square ends. Let's denote the side length of the square ends as "x". The area of the label can be expressed in terms of the box's dimensions. The label covers the four sides of the box, so the total area of the label can be given by: Area = 2 * (length * height + width * height) Since the width is also "x" (the side of the square), the area becomes: Area = 2 * (9 * h + x * h) = 2h(9 + x) Given the area of the label is 72 square metres: 2h(9 + x) = 72. From this, we get: h(9 + x) = 36. Thus, h = 36 / (9 + x). Now, let's relate the height to the side length of the square base. The volume of the box can be calculated as: Volume = length * width * height = 9 * x * h. Substituting h gives: Volume = 9 * x * (36 / (9 + x)). Next, we need to maximize the volume. We can set up the equation for convenience: Volume = (324x) / (9 + x). To find the optimal width x, we would need to take the derivative of the volume function, set it to zero, and solve for x. However, let's use the conditions of the problem directly instead. Using trial and error or analytical methods, we find that when x = 3, h becomes 12 (36 / (9 + 3) = 3). Thus, x = 3 and the height h = 12 yield: Volume = 9 * 3 * 12 = 324 cubic metres. Hence, the volume of the rectangular box is 324 cubic metres.