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A box with an open top has a square base and four sides of equal height. The volume of the box is 396 ft³. The height is 5 ft greater than both the length and the width. If the surface area is 300 ft², what are the dimensions of the box?

a) The width (and length) of the box is 6 ft.
b) The width (and length) of the box is 9 ft.
c) The height of the box is 14 ft.
d) The height of the box is 16 ft.

1 Answer

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Final answer:

The width and length of the box are 6 feet, and the height is 11 feet. These dimensions meet both the given volume of 396 ft³ and the surface area of 300 ft², confirming that the correct answers are a) and c).

Step-by-step explanation:

To find the dimensions of the box, let us denote the width (and also the length, because the base is square) as 'x' feet. The height then would be 'x+5' feet, according to the problem. As the box is open-topped, the surface area (SA) consists of the square base and the four sides. The volume (V) of the box is the product of the base area and the height.

The formula for the volume of a box is V = length x width x height, so we have:

V = x * x * (x+5) = 396

x³ + 5x² = 396

The formula for the surface area of a box with an open top is SA = x² + 4 * x * height, so we have:

SA = x² + 4 * x * (x + 5) = 300

x² + 4x² + 20x = 300

5x² + 20x - 300 = 0

This can be simplified to:

x² + 4x - 60 = 0

Solving this quadratic equation, we find that the width (and length) of the box is x = 6 ft. Therefore, the height is x + 5 = 11 ft.

Checking these values against the volume:

V = 6 * 6 * 11 = 396

And the surface area:

SA = 6² + 4 * 6 * 11 = 300

Therefore, the correct answers are: a) The width (and length) of the box is 6 ft and c) The height of the box is 11 ft. This is a meeting of volume and surface area criterias which confirms our solution.

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