Question

A closed rectangular box whose dimensions are 8 feet by 5 feet by 3 feet has some water in it. When the box is resting on its smallest face, the water is 5 feet deep. How deep will the water be if the box is resting on its largest face? Explain your reasoning.

Answer 1

Answer:

i, 1.875 ft

ii. Since the volume of water is the same when resting on either the smallest or largest face, we we equate the expression for their volume and solve for the depth of water when the closed rectangular box is resting on its largest face.

Step-by-step explanation:

i. How deep will the water be if the box is resting on its largest face?

Since the smallest face is 5 feet by 3 feet (since they are the smallest dimensions), the volume of the 5 feet deep water is V = Ah where A = area of smallest face and h = depth of water.

V = Ah = 5 ft × 3 ft × 5 ft = 75 ft³

This is also the volume of the water when it is resting on its largest face.

Since the dimensions of the largest face are 8 feet by 5 feet (since they are the largest dimensions), then V = A'h' where A' = area of largest face and h' = depth of water when rectangular box is resting on its largest face

So, h' = V/A'

= 75 ft³/ (8 ft × 5 ft)

= 15/8 ft

= 1.875 ft

ii, Explain your reasoning.

Since the volume of the water in the closed rectangular box is always the same, which ever side we turn it to, its volume V = area of face × height. If we write this expression for both the smallest and largest face and equate them, we solve for the depth of the water when it is resting on its largest face, since, we already know the depth of water when it is resting on its smallest face.