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 find the volume of the box, start by determining its dimensions based on the label information. The label, covering the four sides of the box, has an area of 72 square metres. Since the box is 9 metres long and has square top and bottom sides, let the side length of the square top (and bottom) be \(x\) metres. The area covered by the label can be expressed as: \[ \text{Area} = \text{Perimeter of the sides} \times \text{Height} \] The perimeter of the rectangular box (sides only, excluding top and bottom) is: \[ \text{Perimeter} = 2 \times (9 + x) \] The height of the box is \(x\) (since the top and bottom are squares), so the area covered by the label is: \[ 2 \times (9 + x) \times x = 72 \] This simplifies to: \[ 2x(9 + x) = 72 \] \[ x(9 + x) = 36 \] This leads to the quadratic equation: \[ x^2 + 9x - 36 = 0 \] Now we can factor or use the quadratic formula. The solutions to this equation can be found as: \[ x = \frac{-9 \pm \sqrt{9^2 + 4 \times 36}}{2} \] Now simplifying gives: \[ x = \frac{-9 \pm \sqrt{81 + 144}}{2} \] \[ x = \frac{-9 \pm \sqrt{225}}{2} \] \[ x = \frac{-9 \pm 15}{2} \] Calculating the two possible values: 1. \(x = \frac{6}{2} = 3\) 2. \(x = \frac{-24}{2} = -12\) (not viable, since dimensions can't be negative) Thus, \(x = 3\) meters. Now the dimensions of the box are: - Length: 9 m - Width: 3 m - Height: 3 m The volume of the rectangular box is calculated as: \[ \text{Volume} = \text{Length} \times \text{Width} \times \text{Height} = 9 \times 3 \times 3 = 81 \text{ cubic meters} \] Therefore, the volume of the rectangular box is **81 cubic meters**.