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Updated on: 06 Nov 2025, 11:54
Originally posted by Sajjad1994 on 14 Feb 2021, 23:30.
Last edited by Bunuel on 06 Nov 2025, 11:54, edited 3 times in total.
Last edited by Bunuel on 06 Nov 2025, 11:54, edited 3 times in total.
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Volume of
box: 360
box: 360
Volume of
unused space: 90
unused space: 90
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
75% (02:24) correct
25%
(02:10)
wrong
based on 377
sessions
History
Date
Time
Result
Not Attempted Yet
Veritas Prep
Shiny Happy Tube Co. has completed development of a new cylindrical tube and has started taking orders. The tube has a diameter of 6 inches and a height of 10 inches. To ship the tube to customers, the company is designing a rectangular box. The box should be as small as possible in order to use as little material as possible.
In the table below, select a value for the total volume of the smallest box that can be used and an approximate value for the volume of unused space in the box after the tube is placed inside.
Shiny Happy Tube Co. has completed development of a new cylindrical tube and has started taking orders. The tube has a diameter of 6 inches and a height of 10 inches. To ship the tube to customers, the company is designing a rectangular box. The box should be as small as possible in order to use as little material as possible.
In the table below, select a value for the total volume of the smallest box that can be used and an approximate value for the volume of unused space in the box after the tube is placed inside.
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|
Volume of box |
Volume of unused space |
------------ |
| 90 | ||
| 150 | ||
| 210 | ||
| 270 | ||
| 360 |
ShowHide Answer
Official Answer
Volume of
box: 360
box: 360
Volume of
unused space: 90
unused space: 90
rocky620
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15 Feb 2021, 04:57
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Minimum volume of the packing box
Since the diameter of the tube is 6, the length and width of the base will be 6 and the height will be equal to the height of the tube = 10
Volume = 6*6*10 = 360
Unused space in the box after thetube is place inside
Unused space left = Volume of the box - Volume of the tube => 360 - \(\pi * 3^2 *10\) => 360 - 3.14*90 ~ 360 - 270 => 90
Unused space left = 90
Since the diameter of the tube is 6, the length and width of the base will be 6 and the height will be equal to the height of the tube = 10
Volume = 6*6*10 = 360
Unused space in the box after thetube is place inside
Unused space left = Volume of the box - Volume of the tube => 360 - \(\pi * 3^2 *10\) => 360 - 3.14*90 ~ 360 - 270 => 90
Unused space left = 90
03 Jul 2025, 13:47
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Heres how to solve it:
choose the box dimensions:
Box volume = 6 × 6 × 10 = 360 in3.
Compute the cylinder’s volume:
Find the unused space:
choose the box dimensions:
- The tube is 10 in long and has a 6 in diameter, so it’s a cylinder of height 10 and radius 3.
- To fit it in a rectangular box with the least material, you lay it on its side so that the box’s base just contains a 6 in × 6 in square (the tightest possible rectangle that will accommodate a circle of diameter 6).
- Thus the optimal box is 6 in by 6 in by 10 in.
Box volume = 6 × 6 × 10 = 360 in3.
Compute the cylinder’s volume:
- Cylinder volume = π·r2·h = π·(32)·10 = 90 π.
- approximate (π≈3), that is 90·3 = 270 in3.
Find the unused space:
- Unused = box volume – cylinder volume
- 360 – 270 = 90 in3.
sayan640
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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
