Let x be the length of the cut. Then the sides of the box formed are 10-2x by 10-2x by x. (x will be the "height" or depth of the open box; there are two corners for each side so the length of the side is 10-2x)

The volume of...

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Let x be the length of the cut. Then the sides of the box formed are 10-2x by 10-2x by x. (x will be the "height" or depth of the open box; there are two corners for each side so the length of the side is 10-2x)

The volume of the box is l x w x h or:

V=(10-2x)(10-2x)x

You indicate that you are to proceed by guess and check: you should form a table of possible cuts and the associated volume:

Expanding the product we get `V=4x^3-40x^2+100x`

Note that there are some natural restraints on x (the domain of the problem): x must be greater than zero or no box is formed. Also x<5; if x=5 then there is no side left.

x V

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0 0 Not possible but included for completeness

.25 22.5625

.5 40.5

.75 54.1875

1 64

1.25 70.3125

1.5 73.5

1.75 73.9375

2 72

2.25 68.0625

2.5 62.5

2.75 55.6875

3 48

3.25 39.8125

3.5 31.5

3.75 23.4375

4 16

4.25 9.5625

4.5 4.5

4.75 1.1875

5 0 Again impossible but included for completeness

Now the function `V=4x^3-40x^2+100x` has larger values, but they do make sense in the context of the problem.

It appears that the maximum occurs when x is about 2 3/4. You might try some more values in that area.

**It can be shown with calculus that the maximum occurs when `x=5/3` with a maximum volume of `2000/27=74.bar(074)` **

If you are allowed to use technology (a graphing utility or spreadsheet) you can pinpoint the maximum that way also.