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Shiny Happy Tube Co. has completed development of a new cylindrical 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.
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Volume of
box: 360
Volume of
unused space: 90
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45% (medium)

Question Stats:

75% (02:24) correct 25% (02:10) wrong based on 377 sessions

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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
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Volume of
box: 360
Volume of
unused space: 90
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Shiny Happy Tube Co. has completed development of a new cylindrical 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
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Shiny Happy Tube Co. has completed development of a new cylindrical 16 Jun 2024, 12:56
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GMATinsight  KarishmaB chetan2u Would you like to discuss this question ?­
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Shiny Happy Tube Co. has completed development of a new cylindrical 03 Jul 2025, 13:47
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Heres how to solve it:

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.

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