# 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..... 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 width 10-2x

The height is the "folded up" part, which is x

So: our volume is:

`(10-2x)(10-2x)x = 100x-40x^2+x^3`

If we graph this, we get:

But most of this graph doesn't actually make sense, given the physical problem:  You can't cut away squares of length 43 from a length 10 piece of paper.  The length of a corner square must be less than 5, and more than 0, so that is the part of the graph we care about:

So we want to make the length of the corner somewhere in between 1 and 2.

From here, it is a pretty standard calculus problem.  If you're not in calculus, but you have a TI graphing calculator, you can graph it and use the trace button to get a pretty good idea of where that peak point is:  The x value is what you should cut out of the corner, the y value is the volume of the box.

Or, if you want to use trial and error you can try plugging in 1, 2, 1.5, 1.25, 1.375, and try "homing in" on the answer

If you are in calculus, here's what to do:

At that peak point, the derivative is 0 (the slope of the curve is a horizontal line).  So we want to find the place where the derivative is 0.

`y=100x-40x^2+x^3`

`dy/dx = 100-80x+3x^2`

`0=100-80x+3x^2`

`x=-(-80)/(6) +- (sqrt( (-80)^2-4(3)(100) ) )/(6)`

`x=(40)/(3) +- (sqrt(5200))/(6)`

`x=(40)/(3) +- (20 sqrt(13))/(6)`

`x=(40 +- 10 sqrt(13))/(3)`

x=1.3148, or x = 25.3518

But we know that the cut must be between 0 and 5, so we want

x=1.3148

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