What should be the size of the squares cut so that the volume of the open box is maximum? if we take a square sheet of 10cm and cut 4 small squares from the corners and then the paper is folded at...

What should be the size of the squares cut so that the volume of the open box is maximum?

if we take a square sheet of 10cm and cut 4 small squares from the corners and then the paper is folded at the cuts to form an open box

Please explain the whole process of trial and error method with different numbers to find the maximum volume of the open box.....

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mlehuzzah eNotes educator | Certified Educator

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Let's call the length of the squares x.


We need to figure out the length, width, and heigth of the resulting box, and use that to get the volume.


You start with a length of 10, but then you fold up a length of x on each side.
So the length of the resulting box is 10-2x


I can't draw this in this answer box, but here is a picture I found:

http://chestofbooks.com/home-improvement/woodworking/Sloyd-Paper-Cardboard-Iron/Model-No-7-Square-Box.html

scroll down to the grid, and ignore the "No 7"


If the larger square has length 10, and the small corner squares have length x, then the inner box must have length 10-2x.

The same argument goes for the width: the original square has width 10; the resulting box has...

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user2330415 | Student

should the calculus or the trial and error method should be used for the project of class 10. because calculus is in 12th . so which is the safer side. which method to use???

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